Binary Search Trees Part III

This is a discussion on Binary Search Trees Part III within the A Brief History of Cprogramming.com forums, part of the Community Boards category; >You might want to consider #define BST_LEFT 0 and #define BST_RIGHT 1. That would defeat the point of using the ...

  1. #16
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    >You might want to consider #define BST_LEFT 0 and #define BST_RIGHT 1.
    That would defeat the point of using the array. The issue that I addressed was avoiding redundant code by using a boolean test to determine left or right and saving that value in the variable dir. It may look strange at first, but you get used to it quickly.
    You could use the "?" operator, but that has its own problems. For example, you could do int* link = (cond?) &left : &right. A short comment at the definition of link should be enough, I think.

    Singular is datum, plural is data. But according to the rule of common use, either "data is" or "data are" would be acceptable even though the latter is grammatically correct while the former is not.
    You are correct that it is plural of datum. I even looked it up in the webster's dictionary to be sure. But the singular construction may also be correct if you are referring to data as an "abstract mass noun," whatever that is.

    Actually, it is true. The only way to get a degenerate tree is if only one path is available at each step along a search. Barring cases where different degenerate input is mixed together so that the tree is still degenerate, there are only two unique cases where this would happen (ignoring symmetric cases):
    Yes, my counter example was a mix of these two cases.

    Hmm, how about: "However, despite a larger average height, Red Black trees can be more efficient than AVL trees because they spend less time maintaining a balanced structure".
    Looks ok.

    No, a special case is a 2-3-4 tree because it's a specialized B-tree. A Red Black binary search tree is most definitely an abstraction because it is a representation of the 2-3-4 tree using a more comfortable and simpler implementation than that of a B-tree. Maybe an abstraction of a special case?
    I think the 2-3-4 tree is a B-tree with n=4. In that case, 2-3-4 trees are somewhat of an abstraction of B-trees. But are RB trees really an abstraction of the 2-3-4 trees? Aren't they more like a different representation that could be switched if needed? And if both trees convey the same information, abeit differently, then they are not really abstracting each other.(at least in my understanding of abstraction)

    Red Black trees are an abstraction of the infamous B-tree that is generally used for database applications. This abstraction has largely been forgotten because Red Black trees are believed to be simpler without it, but I feel that knowing the rationale behind the red black scheme helps to understand the concept. Red Black trees are not as optimal as AVL trees; they have a maximum height of 2 * log N. This is not necessarily a bad thing because Red Black trees do not work nearly as hard to maintain the strucutre. Thus, a Red Black tree may perform better than an AVL tree despite having a larger average height. Red Black trees are best suited to applications where the input is largely random with occasional spouts of ordered data.
    I kind of like the "not necessarily a bad thing". Maybe even you could try something like "Not necessarily a bad thing: Red Black trees do not work nearly as hard to maintain their tree structure." I've written another way below.

    Red Black trees are not as asymtotically efficient as AVL trees; they have a maximum height of 2 * log N. But they do have their advantages: Red Black trees do not work nearly as hard maintain their structure. Thus, a Red Black tree, with larger average height, may perform faster than an AVL tree. And if the data is random with occasional spouts of ordered data, choose a Red Black tree; reduced operation costs may more than make up for an increased tree size.

    I've changed a few words here and there. Nothing too major.
    Last edited by okinrus; 10-02-2004 at 02:53 PM.

  2. #17
    Code Goddess Prelude's Avatar
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    Okay, here's the final draft that I've asked Hammer to add to the FAQ:


    Part III - Balanced Binary Search Trees


    Binary search trees are easy to implement and understand for beginning programmers. Unfortunately, in a production environment these trees are typically avoided for one simple reason: performance in the worst case. Let us assume that we have a basic recursive binary search tree with a pretty print routine:
    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    struct node {
      int data;
      struct node *link[2];
    };
    
    struct node *make_node ( int data )
    {
      struct node *tree = malloc ( sizeof *tree );
    
      if ( tree == NULL )
        return NULL;
    
      tree->data = data;
      tree->link[0] = tree->link[1] = NULL;
    
      return tree;
    }
    
    struct node *insert_node ( struct node *tree, int data )
    {
      if ( tree == NULL )
        tree = make_node ( data );
      else {
        int dir = data > tree->data;
    
        tree->link[dir] = insert_node ( tree->link[dir], data );
      }
    
      return tree;
    }
    
    void tree_structure ( struct node *tree, int level )
    {
      int i;
    
      if ( tree == NULL ) {
        for ( i = 0; i < level; i++ )
          printf ( "\t" );
        printf ( "~\n" );
    
        return;
      }
    
      tree_structure ( tree->link[1], level + 1 );
    
      for ( i = 0; i < level; i++ )
        printf ( "\t" );
      printf ( "%d\n", tree->data );
    
      tree_structure ( tree->link[0], level + 1 );
    }
    For the most part, this simple tree is more than enough. Granted, one may need to add more features such as deletion, searching, and possibly some symbol table operations, but as binary search trees go, this one is pretty good. If you have any trouble at all understanding the code just given, please refer to parts I and II of this tutorial. It only gets harder, and a sound understanding of the basics is critical. :-)


    Degenerate Trees

    Any operation that involves searching on a binary search tree will typically perform in logarithmic time on average. On average this is good, and binary search trees serve their purpose wonderfully. However, problems arise when the average case turns into the worst case. Many algorithms with a logarithmic average case suffer from quadratic worst case performance, and basic binary search trees are no exception. If the data are inserted into a basic binary search tree in such a way that there is only one way to go at each node, the result will be a tree with a height equal to the number of elements in the tree.

    As you learned earlier in this series, the height of a tree is the longest path from the root to a leaf. When the height is short, the tree branches nicely and searching is performed in logarithmic time:
    Code:
           8
    
      4        12
    
    1   6    9    13
    However, if the height is not short then the tree begins to take on the appearance of a linked list rather than a binary search tree:
    Code:
    1
    
      4
    
        6
    
          8
    
            9
    
              12
    
                13
    This is called the worst case of a binary search tree, and it is known as a degenerate tree because performance is reduced drastically. For this reason, basic binary search trees aren't as useful as one might think at first.

    Fortunately, all is not lost. There are many ways to force a binary search tree to maintain a short height even when faced with certain bad input. Unfortunately, these methods are generally quite complex and very few programmers enjoy writing or maintaining them. Luckily for you, I am one of those who enjoy it very much. :-)

    In this part of the binary search tree tutorial, we will cover balanced binary search trees. These trees use various clever techniques that force a tree to branch evenly left and right. Because of the vast number of techniques for balancing a binary search tree, I cannot hope to fit all of them into a single article. Therefore, I will focus on three popular balanced trees: AVL trees, Red Black trees, and Splay trees.


    Guaranteed Logarithmic Performance

    Balanced binary search trees are designed with good performance in mind, regardless of the ordering of insertions and deletions. Balanced trees fall into one of three categories. The first we have already covered in part II, Randomized trees. These trees reduce the chances of a degenerate tree to virtually nothing by utilizing random numbers in operations that modify the tree structure. Randomized algorithms are becoming more popular because they are generally simple to write and maintain as well as extend. Their performance is rather difficult to analyze, but most of us just leave that to the brainy mathematicians and trust that the algorithms work properly. :-)

    The second category is by far the most common. Optimized algorithms guarantee good performance by working very hard to maintain the best possible case without hurting performance. Optimized algorithms are difficult to write and maintain, and a PHD in algorithm analysis is usually required to extend them without breaking something. ;-) The optimized balanced tree algorithms that we will be studying in this article are for AVL trees and Red Black trees.

    AVL trees are guaranteed to perform well because they are balanced according to height. If the height of a node's subtree grows higher than the other by more than one, the tree is restructured to fix it. Most of the time, the height of an AVL tree will be very close to log N, with a worst case of 1.44log N. This is as close to optimal as possible while still being practical. AVL trees are best used when the data are known to come in sorted order with high frequency, and any searches will be fairly random.

    Red Black trees are an abstraction of the infamous B-tree that is generally used for database applications. This abstraction has largely been forgotten because Red Black trees are believed to be simpler without it, but I feel that knowing the rationale behind the red black scheme helps to understand the concept. Red Black trees are not as optimal as AVL trees; they have a maximum height of 2 * log N. However, because Red Black trees do not work as hard as AVL trees to maintain a balanced structure, they tend to be more efficient in many applications, despite having a larger average height. Red Black trees are best suited to applications where the input is largely random with occasional spouts of ordered data.

    The third category consists of Amortized algorithms. These are algorithms that guarantee performance over a large number of operations. The Splay tree is what we will be covering. A splay tree requires that every time an item is searched for, it is lifted to the root of the tree. Splaying is not like the root insertion routine we wrote in Part I though; there are a few slippery areas. Most notable about splay trees is the extreme difficulty in analyzing them. This is because the tree structure is dynamic in ways that cannot be predicted since the splaying is performed based on search hits. While any one operation on a splay tree may be inefficient, it has been proven (through mathematical proofs that I will never understand) that the total running time of all operations put together is logarithmic. Hence the title of Amortized; the expensive operations are averaged with the more abundant less expensive operations to give excellent performance overall. Splay trees are best suited to applications where data is searched for in clusters or the same few items are searched for often.



    AVL Trees


    Concept

    The AVL tree is named after its creators, G.M. Adelson-Velsky and E.M. Landis. The general idea is that for every node, the height of its left subtree never differs by more than 1 by the height of its right subtree. When all nodes in a tree meet this requirement, the tree is balanced. The following is a diagram showing an AVL tree of ten nodes:
    Code:
                            8\
    
              4|                        12|
    
         2|       6|            10|           14\
    
       a    b   c    d      9|      11|     i     15|
    
                          e    f  g     h       j     k
    Notice that each item has a special character added. In a non-ASCII world I would give you graphical diagrams, but for now these will do. For each node, the number is the data item and the character is what is called a balance factor. | means that the left and right subtrees are of equal height, \ means that the right subtree is taller than the left subtree, and / means that the left subtree is taller than the right subtree. Lower case letters designate null items where a new item may be placed.

    Note that when some new items are added, the tree does not go out of balance. If we added a node a, b, c, d, or i then no restructuring is necessary. If a new item is added anywhere else, the structure of the tree will need to be adjusted to account for it.

    The tree structure will need to be fixed when a node with a balance factor of \ has its right subtree grow in height, or when a node with a balance factor of / has its left subtree grow in height. It isn't too difficult to see that the problem areas arise in four distinct cases, two for a right imbalance and two for a left imbalance. The first case is where a node's right subtree whose right subtree has grown in height. In the following diagram, A's right node B has increased its right height:
    Code:
        A\
    
    h        B\
    
           h    h+1
    This case can be fixed with a single rotation left for A:
    Code:
           B|
    
      A|       h+1
    
    h    h
    Also notice that the balance factors for both A and B have been modified to reflect the structural change. The biggest complexity with AVL trees is maintaining these balance factors correctly.

    The second case consists of a node whose right subtree has had its left subtree grow in height. In the following diagram, A's right node B has increased its left height through C:
    Code:
        A\
    
    h         B/
    
           C     h
    
       h-1    h
    The solution to this case is slightly more complex than the first case. Two rotations are required to restore a balanced structure. First, a right rotation at B and then a left rotation at C:
    Code:
        A                                 C|
    
    h         C                     A            B
                        ---->
          h-1    B               h     h      h     h
    
               h    h
    Notice that I've neglected to mention what the balance factors are for C in the first step and for A and B in the last step. The intermediate step doesn't really matter because modification of the balance factors will take place either before the first rotation or after the second rotation. The step was included only for illustrative purposes. The reason I have omitted the balance factors on some of the key nodes is because a double rotation requires a fairly nasty modification of balance factors that depends heavily on the current setting of C. When we get to the code you'll see what I mean.

    The symmetric cases (where the imbalance is in A's left subtree) are identical. In our implementation we will take advantage of this to make the code shorter and easier to follow.

    Removal of an item is more complicated and can be performed several ways. By far the simplest is to use simple deletion by copying (as explained in Part I). However, the algorithm cannot end there because an imbalance may have occurred by shortening a node's subtree to the point where rebalancing is required. Therefore, the algorithm must then move down to the successor (or predecessor, depending on how the deletion is implemented) and walk back up the tree from there while changing balance factors and performing rotations if necessary.

    For node removal the principles are the same, consisting of single and double rotations with careful modification of the balance factors, so we will not cover the process until later. When we get to the implementation for remove_node, hopefully will become clear. :-)


    Exercises

    1) Diagram the symmetric cases for rebalancing, where A's left subtree has increased in height.

    2) Using the descriptions given, try to work out what changes need to be made to the balance factors of A and C when a double rotation is performed.

    3) What is the time complexity of insertion and removal from an AVL tree?



    Balance Factors

    AVL trees theoretically only need two extra bits added to a node to handle the balance factors. This is the case when the balance factors are only allowed to hold three values, -1 when the left subtree is higher, +1 when the right subtree is higher, and 0 when both subtrees are of equal height. In our implementation these values will be slightly different, where 0 is used for a higher left subtree, 1 for a higher right subtree, and 2 for equal height. The choice of value for equal height for our implementation is arbitrary as long as it isn't either 0 or 1.

    The C structure for an AVL node can be written a number of ways, but the most common are using a signed char for the balance factor:
    Code:
    struct node {
      int data;
      signed char balance;
      struct node *link[2];
    };
    Using a signed integer:
    Code:
    struct node {
      int data;
      int balance;
      struct node *link[2];
    };
    And sometimes a bitfield is used with the pretense of saving space:
    Code:
    struct node {
      int data;
      int balance: 2;
      struct node *link[2];
    };
    The space savings from this structure are usually false as the space for a full integer will be used for alignment purposes. To get the space savings, the fields of the structure would need to be ordered in an implementation-dependent manner. Therefore, the use of a bitfield is not very portable. Our implementation will use a signed integer for simplicity, but it is recommended to use the smallest possible type for the balance field because it does not need to hold more than three values.

    Some implementations of AVL trees will write the balance factors as weight counts. A weight count is the number of nodes in both subtrees. The AVL scheme can be maintained by taking the difference of the weight count for the left and right subtree and rebalancing if they differ by more than 1. This often results in less convoluted code than the strict -1, 0, +1 balance factor scheme, but some slippery areas require hackish feeling solutions and the trees can only grow so large before a larger data type is required for the weight count. As an example, consider an implementation with 16 bit integers where the node structure looks like this:
    Code:
    struct node {
      int data;
      int balance;
      struct node *link[2];
    };
    The tree could only have a total log2(N) nodes where N is the maximum value that 16 bits could hold, thus, 32,767 nodes. While that is a large tree, in a realistic environment using an AVL tree, one can expect larger trees by at least an order of magnitude. Therefore the weight count implementation is limited by the data type of the balance factor.


    Exercises

    1) What other ways can you think of to represent balance factors?

    2) Determine the most space efficient order for the fields in an AVL node using a bitfield for the balance factor.

    3) Why are two bits the minimum for the AVL balance factor?

    4) Try to guess why our implementation needs 0 and 1 for the left and right balance factors and why the equal height balance factor can be anything.



    Insertion

    Insertion into an AVL tree consists of three parts: Searching for any empty location to insert at, inserting a new node at that location, and rebalancing the tree. The search and insertion steps are identical to that of a basic binary search tree. However, before we go any further a bit of preliminary code is needed. First we need to modify the basic binary search tree node to allow for balance factors. This is a simple addition of the balance field:
    Code:
    struct node {
      int data;
      int balance;
      struct node *link[2];
    };
    This field will only hold three values: 0, 1, and 2. Each value represents a balance factor for a higher left subtree, a higher right subtree, and both subtrees of equal height, respectively. The reason that 0 and 1 are used rather than -1 and +1 is so that we can use the boolean dir variable (as seen in parts I and II) to eliminate the redundant code of symmetric cases.

    Another auxiliary function that we might find useful is a function designed to create a node:
    Code:
    #include <stdlib.h>
    
    struct node *make_node ( int data )
    {
      struct node *tree = malloc ( sizeof *tree );
    
      if ( tree == NULL )
        return NULL;
    
      tree->data = data;
      tree->balance = 2;
      tree->link[0] = tree->link[1] = NULL;
    
      return tree;
    }
    Note that new nodes are always balanced with equal height because both the left and right subtrees are empty. While make_node returns a null pointer if a memory allocation error occurs, our insertion algorithm will ignore it and assume that the new node was successfully added for simplicity sake. For production quality implementations, this should be one of the first issues to be addressed.

    Next, because it is a good habit to modularize commonly used operations we will write separate single_rotate and double_rotate functions. These functions will be used heavily in all balanced trees, so it would be a good idea to write them now as they do nothing special with AVL balance factors:
    Code:
    struct node *single_rotate ( struct node *tree, int dir )
    {
      struct node *save = tree->link[dir];
    
      tree->link[dir] = save->link[!dir];
      save->link[!dir] = tree;
    
      return save;
    }
    
    struct node *double_rotate ( struct node *tree, int dir )
    {
      struct node *save = tree->link[dir]->link[!dir];
    
      tree->link[dir]->link[!dir] = save->link[dir];
      save->link[dir] = tree->link[dir];
      tree->link[dir] = save;
    
      save = tree->link[dir];
      tree->link[dir] = save->link[!dir];
      save->link[!dir] = tree;
    
      return save;
    }
    Now, it is certainly possible to add modification of the balance factors to the rotation code, and many implementations do so. However, this adds unnecessary complexity because changing balance factors is different for insertion and removal, and it is better to keep the modification of balance factors localized and separate from the structural changes. An additional advantage is that we can reuse these two functions for other balanced trees without changing them. :-)

    Insertion into an AVL tree, as said previously, is identical to that of a basic binary search tree with the exception of the structural changes and maintenance of balance factors on the walk back up to the root. The only issue is determining when rebalancing is required. If we rebalance at every step up the tree, the AVL condition will not be properly maintained without a slew of special cases, so it makes sense to avoid rebalancing if a node is not inserted such as if it already exists in the tree (we will not implement this feature) or after rotations have completely fixed the structure of the tree. At this point no further rotations or balance factor changes are needed, so we need a way to notify the algorithm when (and especially when not) to rebalance itself.

    The solution is a status flag passed to the insertion function. Assuming that we have a function to handle rebalancing, a recursive implementation is immediate:
    Code:
    struct node *insert_node ( struct node *tree, int data, int *deeper )
    {
      if ( tree == NULL ) {
        tree = make_node ( data );
        *deeper = 1;
      }
      else {
        int dir = data > tree->data;
    
        tree->link[dir] = insert_node ( tree->link[dir], data, deeper );
    
        if ( *deeper )
          tree = rebalance_insert ( tree, dir, deeper );
      }
    
      return tree;
    }
    The deeper status flag is true if the tree has grown in height and false otherwise. If deeper is true then the node enters a rebalancing routine that changes balance factors and rotates as necessary. This code is painfully simple and shocking to everyone who has seen an AVL insertion algorithm before because it is so short. :-) Part of this is the modularization of deferring work to rebalance_insert and part of it is the use of recursion to save ourselves the trouble of either maintaining a stack or using a clever walk down scheme to avoid moving back up the tree.

    Using a stack instead of recursion is a viable alternative, but it buys us little with a balanced tree. There are some performance issues, but in this case clarity wins out over speed. Knuth describes a way to insert into an AVL tree without using a stack or recursion by saving the point at which the tree might need to be rebalanced at each insertion because AVL insertion requires at most either a single or a double rotation. This is an excellent idea and our implementation would have used it if not for the desire to give insert and deletion a similar feel. Insertion requires at most one rotation, but deletion could require up to log2(N) rotations. Because of that, the iterative deletion algorithm must use a stack of both visited nodes and directions taken. Our implementation uses recursion for both insertion and deletion, and the result is a similar solution in both cases. I believe that this will aid in understanding.

    The rebalance_insert function handles the two cases described in the AVL tree concept section. The dir variable is used to avoid the symmetric cases by merging them into one. If dir is 0 then a left rebalance is performed, otherwise dir is 1 and a right rebalance is performed. Nifty, isn't it? :-)
    Code:
    struct node *rebalance_insert ( struct node *tree, int dir, int *new_depth )
    {
      if ( tree->balance == dir ) {
        if ( tree->link[dir]->balance == dir ) {
          tree->link[dir]->balance = 2;
          tree->balance = 2;
    
          tree = single_rotate ( tree, dir );
        }
        else {
          fix_balance_factors ( tree, dir );
    
          tree = double_rotate ( tree, dir );
        }
    
        *new_depth = 0;
      }
      else if ( tree->balance == !dir ) {
        tree->balance = 2;
        *new_depth = 0;
      }
      else
        tree->balance = dir;
    
      return tree;
    }
    This is where the complexity of the AVL tree shows up. There are three cases to consider with the tree balance. If the balance is equal then we need to propagate the fact that the subtree has increased in height. This case requires a simple change to the balance factor showing that the dir (0 == left, 1 == right) subtree has grown deeper. The status flag remains true because we still do not know if the change in height has cause an imbalance, so the next node in the path back up the tree will call rebalance_insert again until a case where a rotation is performed, or it is determined that a rotation is not necessary by the recursive calls all returning without a rotation.

    The second simplest case is when the node's shorter subtree has grown higher. In this case the tree has gotten more balanced and no rotations are necessary, so the balance factor is set to reflect equal balance and the status flag is cleared. Rebalancing is done for this insertion.

    The third case is when the node's taller subtree has grown higher. The only solution at this point is a rotation, and we now enter the two cases described in the concept section. If the node's dir link has a balance factor of dir then Case 1 applies and a single rotation is performed. The balance factors of the node and its dir link are change to show that the tree has now become balanced. Both balance factors are of equal height as the diagram in the concept section shows.

    Case 2 applies if the node's dir link has a balance factor of equal or !dir. According to the diagram, A is tree, B is tree->link[dir], and C is tree->link[dir]->link[!dir]. This means that either of C's subtrees have grown higher, or C is the new node. We know that a double rotation is needed, but how to change the balance factors before the rotation remains a mystery because you already know that it depends on the present setting of the balance factors of C. If C's balance factor is dir then A's new balance factor is 2, signifying equal height and B's new balance factor is !dir. If C's balance factor is !dir then A's new balance factor is dir and B's new balance factor is 2. If C's balance factor is 2 then both A and B are given balance factors of 2 as well. After all of these cases, we make sure that C has a balance factor of 2 (remember the diagram).

    The code for this magic is simple even if exactly why those balance factors are needed is not:
    Code:
    void fix_balance_factors ( struct node *tree, int dir )
    {
      if ( tree->link[dir]->link[!dir]->balance == dir ) {
        tree->balance = 2;
        tree->link[dir]->balance = !dir;
      }
      else if ( tree->link[dir]->link[!dir]->balance == !dir ) {
        tree->balance = dir;
        tree->link[dir]->balance = 2;
      }
      else
        tree->balance = tree->link[dir]->balance = 2;
    
      tree->link[dir]->link[!dir]->balance = 2;
    }
    That is all there is to it! Three simple routines and we now have a working implementation for AVL insertion. It is highly recommended that you compile and run this code. Step through it in a debugger and try to understand what it is doing and why. Nothing is better than that kind of hands on fiddling for understanding an advanced data structure.


    Exercises

    1) fix_balance_factors maintains balance factors based on the balance factor of C. Why are these particular changes necessary? In what situations will each case be used? Will different balance factors still work? Which ones?

    2) Write down a test run for various inputs on a piece of paper. Follow the code exactly and see what it does. Does the algorithm do what you expected?

    3) Can this insertion algorithm be rewritten without the status flag?

    4) Rewrite insert_node without recursion using what you learned in Part I. Is it more efficient? Is the performance gain worth the effort?

    5) Rewrite insert_node without recursion or the use of a stack. Hint: Save the first node along the search path without an equal balance factor. This is the point where rebalancing may be needed.



    Deletion

    Removing a node from an AVL tree is more complicated than inserting a node. The problem is removing an internal node without causing an imbalance elsewhere in the tree that is not known. There are a few different ways of going about AVL deletion, but we will be using a recursive scheme similar to the insertion algorithm.

    Before we get started, we need a helper function, just like with insertion.
    Code:
    void destroy_node ( struct node *tree )
    {
      free ( tree );
    }
    destroy_node is a placeholder for more complex nodes that may require releasing their own memory. If the node is released before that happens then we have a memory leak.

    Like insert_node, remove_node will also make use of a status flag for the same reasons. We would not want to perform rotations or change balance factors without reason because that would ruin the already fragile AVL data structure. This is why AVL trees are seen as difficult, even the smallest change can destroy the algorithm because it relies so heavily on the proper setting of balance factors.

    As we did in Part I, we will use a deletion-by-copying algorithm, but unlike in Part I we will do it almost completely with recursion. The algorithm without any balancing mumbo jumbo looks like this:
    Code:
    struct node *remove_node ( struct node *tree, int data )
    {
      if ( tree == NULL )
        ;
      else if ( data == tree->data ) {
        if ( tree->link[0] != NULL && tree->link[1] != NULL ) {
          struct node *heir = tree->link[0];
    
          while ( heir->link[1] != NULL )
            heir = heir->link[1];
    
          tree->data = heir->data;
    
          tree->link[0] = remove_node ( tree->link[0], tree->data );
        }
        else {
          struct node *save = tree;
    
          tree = tree->link[tree->link[0] == NULL];
    
          destroy_node ( save );
        }
      }
      else {
        int dir = data > tree->data;
    
        tree->link[dir] = remove_node ( tree->link[dir], data );
      }
    
      return tree;
    }
    If the item is not found then you can ignore it as above or set up some kind of notification framework. If the item is found then the inorder predecessor is found and its data is copied into the item to delete. Then, and this is where we do something different, remove_node is called recursively on the node's left subtree with the inorder predecessor's data as the key. Now, what was the deletion of data originally becomes the deletion of that node's predecessor. This trick is used to turn the difficult case where the node to be deleted has two children into one of the easy cases where the node to be deleted only has one or zero children. Read this code carefully and make sure you understand how it works because we are about to add balancing to the equation.

    To convert remove_node into an algorithm that can remove an item from an AVL tree and still maintain the AVL condition, we need to consider only two things: how to figure out when to rebalance and where to rebalance. The first question is simple because we can use the same solution as last time, use a status flag. The second problem is not as simple as insertion because we have two unique paths that may need rebalancing. Remember that deletion from an AVL tree may take up to log2(N) rotations, so we need to consider that every node from the predecessor to the root could need rebalancing. This means that we must add a rebalancing check for both the data path and the predecessor path.

    The second part to this question is where do we set the flag to true and where to false? Setting it to false is simple, the rebalance function will handle that and if the item is not found the remove_node should make sure that no rebalancing is performed. Where to set the flag to true is more difficult. At what point do we know for a fact that the tree has gotten shorter?

    When we actually remove a node, of course. Just like insertion where when we inserted a node we assumed that the tree grew in height, we can assume that the tree shrinks in height when we actually remove a node. Now, the only time we actually remove a node is when the predecessor is replaced with one of its children. Remember that deletion by copying does not actually remove the first node, it only replaces its data.

    With all of that in mind, we can come up with something workable on the assumption that rebalance_remove will handle the nitty-gritty stuff:
    Code:
    struct node *remove_node ( struct node *tree, int data, int *shallower )
    {
      if ( tree == NULL )
        *shallower = 0;
      else if ( data == tree->data ) {
        if ( tree->link[0] != NULL && tree->link[1] != NULL ) {
          struct node *heir = tree->link[0];
    
          while ( heir->link[1] != NULL )
            heir = heir->link[1];
    
          tree->data = heir->data;
    
          tree->link[0] = remove_node ( tree->link[0], tree->data, shallower );
    
          if ( *shallower )
            tree = rebalance_remove ( tree, 0, shallower );
        }
        else {
          struct node *save = tree;
    
          tree = tree->link[tree->link[0] == NULL];
          destroy_node ( save );
    
          *shallower = 1;
        }
      }
      else {
        int dir = data > tree->data;
    
        tree->link[dir] = remove_node ( tree->link[dir], data, shallower );
    
        if ( *shallower )
          tree = rebalance_remove ( tree, dir, shallower );
      }
    
      return tree;
    }
    Notice that the first call to rebalance_remove in the two child deletion case uses 0 instead of dir for the direction. This is because we already know what direction we are going along the path, so dir is not needed. The status flag shallower is set to false if the item is not found and true if the item is found, but only after the predecessor is actually removed from the tree. The algorithm should be obvious by now, so we will look at rebalance_remove:
    Code:
    struct node *rebalance_remove ( struct node *tree, int dir, int *new_depth )
    {
      if ( tree->balance == dir )
        tree->balance = 2;
      else if ( tree->balance == !dir ) {
        if ( tree->link[!dir]->balance == !dir ) {
          tree->link[!dir]->balance = 2;
          tree->balance = 2;
    
          tree = single_rotate ( tree, !dir );
        }
        else if ( tree->link[!dir]->balance == 2 ) {
          tree->link[!dir]->balance = dir;
    
          tree = single_rotate ( tree, !dir );
        }
        else {
          fix_balance_factors ( tree, !dir );
    
          tree = double_rotate ( tree, !dir );
        }
      }
      else {
        tree->balance = !dir;
        *new_depth = 0;
      }
    
      return tree;
    }
    Rebalancing in AVL deletion is not as simple as insertion. As with insertion, there are three cases with the balance factor of A. If A's balance factor is dir then the taller subtree has grown shorter and we reset the balance factor to 2. This change does not rebalance the tree, so we do not clear the status flag and allow rebalancing to propagate upward. If A's balance factor is 2 then we reset A's balance factor to show that the other subtree has gotten taller. This change tells us that no more rebalancing is required, so the flag is cleared and life is good.

    The case where A's balance factor is !dir means that the shorter subtree has grown shorter and rotations are required. At this point there are three cases for rebalancing. If B's balance factor is also !dir then the familiar single rotation is performed and balance factors for A and B become 2. A new case where B's balance factor is 2 shows up. A single rotation will restore the AVL condition, but setting the balance factors of A and B to 2 will not work. In this case, only B's balance factor is changed to dir before the rotation is made.

    The final case, where B's balance factor is dir, requires a double rotation and a call to fix_balance_factors. The good news is that this case is identical to the equivalent case in insertion, so we can reuse fix_balance_factors just as we reused single_rotate and double_rotate.

    After these three cases, we do not reset the status flag because a rotation does not necessarily restore balance to an AVL tree during deletion. We will continue to propagate the rebalancing upward until a case is encountered where we know we can stop, or we have reached the root node and remove_node terminates.

    This is all that is required for AVL deletion. When you look at it the right way, the algorithm is very simple. Most of the simplicity comes from modularization, reuse, and the use of recursion. A non-recursive version of the algorithm requires a stack of both nodes along the path and directions taken, so a truly elegant non-recursive deletion algorithm isn't possible as it is with insertion.


    Exercises

    1) Only one case in rebalance_remove clears the status flag, why does that case mean that rebalancing can stop?

    2) AVL deletion adds a third rotation case where B's balance factor is 2, why must that case be treated separately from the usual single rotation case?

    3) Work through several runs of input on paper using remove_node's algorithm as a guide. Does it work like you expected?

    4) Rewrite remove_delete without using recursion. Is it simpler? Is it worth the effort?



    Red Black Trees


    Concept

    While Red Black trees are a data structure of their own these days, they were originally an abstraction. I feel that people have trouble with Red Black trees because the abstraction has largely been dismissed and the rationale behind the rules has been forgotten. So we will cover the underlying data structure that the Red Black tree is an abstraction of and then we will put it aside in favor of an implementation simply following the rules. This way not only is the data structure understood, but the meaning behind it is as well.

    A B-tree of order 4, also called a 2-3-4 tree or 2-4 tree, is a search tree with nodes that can have either 2, 3, or 4 links to other nodes, called (unimaginatively) 2-nodes, 3-nodes, and 4-nodes, respectively. Following this line of thinking, a binary tree consists solely of 2-nodes.
    Code:
       2-node             3-node             4-node
    
         1                 1 2                1 2 3
    
      ~     ~            ~  ~  ~             ~ ~ ~ ~
    A B-tree is kept balanced by guaranteeing that all leaf nodes are the same distance from the root. This is done by filling nodes until they become 4-nodes and then splitting them upward. When a new item is added to a 4-node, it needs to be split into three 2-nodes before the insertion can be made.
    Code:
    Insert 4:
    
      1 2 3                  2                   2
                --->                   --->
     a b c d             1       3           1       3 4
    
                       a   b   c   d       a   b    c d e
    The benefit of a data structure like this is that the height of the tree remains relatively small. Also, the advantage of keeping all leaves at the same distance from the root means that the tree is always perfectly balanced.

    Unfortunately, the code for a B-tree is very specialized. We would like to be able to use the familiar binary search tree operations rather than writing all new operations for a B-tree. That's where the Red Black abstraction comes in. Rather than having explicit 2, 3, and 4-nodes, we can use only 2-nodes (thus a binary tree) and use a flag to tell us if the link is a logical link (making it a part of a 3 or 4-node), or a hard link (making it a link to the next 2, 3, or 4-node). The current accepted terminology gives logical links and hard links a different color, with logical links being red and hard links being black.

    The color of a node (R == red, B == black) follows the item.
    Code:
       2-node                  3-node                      4-node
    
         1B             1B                2B                 2B
    
       ~    ~       ~      2R    or    1R      ~        1R         3R
    
                         ~    ~      ~    ~           ~    ~     ~    ~
    Because a 3-node can be represented one of two ways with two 2-nodes, we need to consider both. All of these representations are common and well understood. The only problem is coming up with rules that meet the requirements of a B-tree of order 4.

    Rule 1: Every path in a Red Black tree must have the same number of black links.

    This makes sense because for the B-tree condition to be met, we need a way to ensure that all leaves are the same distance from the root. Because black links are "real" links in the abstraction and red links are "fake" links, we can ignore the red links and just make sure that every path in a Red Black tree has the same number of black links. This meets the requirement that all leaves in a B-tree are the same distance from the root.

    Rule 2: Two red links cannot appear consecutively on any path in a Red Black tree.

    This also makes sense when you consider that red links make 3-nodes and 4-nodes. Because the largest node we can use is a 4-node, and a 4-node is represented by a black node with two red links, having two red links consecutively would signify the need to either fix a broken 4-node, or perform a split. Either way the B-tree condition is violated and needs to be fixed.

    Rule 3: The root must be black.

    This rule is not necessary with general Red Black trees, but it is usually adhered to because it simplifies the algorithms. We will use it as a hard rule because it does not make much sense to allow a red root when using the B-tree abstraction.

    Tossing aside the abstraction, we have three rules that enforce the Red Black tree condition and help simplify the algorithms:

    Rule 1: Every path in a Red Black tree must have the same number of black nodes.
    Rule 2: Two red links cannot appear consecutively on any path in a Red Black tree.
    Rule 3: The root must be black.

    By following these rules in an implementation, we can guarantee logarithmic search times.


    Exercises

    1) Why would a red root make some algorithms more difficult?

    2) Devise as many cases as you can where two consecutive links would be red, and then explain why they are a violation of the B-tree condition.



    Link Coloration

    In a Red Black tree, the color of a node is the balance factor, much like AVL trees. Because it is more difficult to actually give links a color, we will be giving each node a color. This can be done numerous ways, but the most common are an enumeration:
    Code:
    enum rb_color { RED, BLACK };
    
    struct node {
      int data;
      enum rb_color color;
      struct node *link[2];
    };
    And a flag that is true if the node is red and false if it is black:
    Code:
    struct node {
      int data;
      int red;
      struct node *link[2];
    };
    Because the flag option only requires one extra bit in theory, and it is possible to remove even that requirement, Red Black trees can be more space efficient than AVL trees. Concerning the data type of the balance factor for Red Black trees, all comments from the AVL balance factor section apply.


    Exercises

    1) Try to come up with as many ways to represent link coloration as you can.

    2) Consider ways that the link coloration could be removed completely. Hint: Swap left and right links and test if the left link is greater than the right link.



    Insertion

    Insertion into a Red Black tree is fairly straightforward. The algorithm we will use is strikingly similar to the insertion algorithm for AVL trees. However, as for AVL trees we need to define some functions to help us first. We will start with the structure declaration for a Red Black node, and functions to create and destroy a Red Black node:
    Code:
    struct node {
      int data;
      int red;
      struct node *link[2];
    };
    
    struct node *make_node ( int data )
    {
      struct node *tree = malloc ( sizeof *tree );
    
      if ( tree == NULL )
        return NULL;
    
      tree->data = data;
      tree->red = 1;
      tree->link[0] = tree->link[1] = NULL;
    
      return tree;
    }
    
    void destroy_node ( struct node *tree )
    {
      free ( tree );
    }
    From the link coloration section, you know that I have chosen to use a flag that is true if the node is red and false if the node is black. But why is a new node colored red? The answer is simple, if a new node were colored black then this would increase the black height of the tree, thus violating rule 1.

    Naturally, if the parent of the new node is red then we have violated rule 2 of the Red Black specification. Rather than try to watch for and fix violations of both rule 1 and rule 2, insertion will be careful to only violate and fix rule 2. Deletions will suffer from having to handle violations of rule 1 as well, but we will cross that bridge when we get to it.

    The next helper function tests a node to see if it is black. Because we will be treating null pointers as black nodes to avoid violations of rule 2, we need to check if the node is null before dereferencing it. The code is simple and removes a bit of clutter from the rebalancing algorithms:
    Code:
    int is_black ( struct node *tree )
    {
      return tree == NULL || !tree->red;
    }
    Now we come to the insertion function itself, insert_node. Unlike the insertion function for AVL trees, we need an auxiliary function to call the recursive function that does the real work. The reason for this need is to keep from violating rule 3 by setting the root to be black after each insertion. The good news is that we do not need a status flag to tell us if the height of the tree has grown, so we end up with this for insert_node:
    Code:
    struct node *insert_node ( struct node *tree, int data )
    {
      tree = insert_node_r ( tree, data );
    
      if ( tree->red )
        tree->red = 0;
    
      return tree;
    }
    Just what you would expect, but I imagine that you are feeling a sense of foreboding. What could insert_node_r possibly be like? Is it a hideous piece of code that you'll never hope to understand? I sure hope not if you've gotten this far. :-)
    Code:
    struct node *insert_node_r ( struct node *tree, int data )
    {
      if ( tree == NULL )
        tree = make_node ( data );
      else {
        int dir = data > tree->data;
    
        tree->link[dir] = insert_node_r ( tree->link[dir], data );
    
        tree = rebalance_insert ( tree, dir );
      }
    
      return tree;
    }
    This is nothing new, by now you should be familiar with all of the techniques used and you may now marvel at how concise it is. Sadly, rebalance_insert is not quite as elegant.

    Rebalancing in a Red Black tree is required if a red node has a red child. There are three possible outcomes depending on the orientation of the child and the color of the node's sibling. Here are the cases (ignoring symmetric cases):
    Code:
    Case 1:
    
       1B                    2B
    
    0B    2R     --->     1R    3R
    
             3R        0B
    In this case, the rule 2 violation is at 2. Because 0 is black, we have no choice but to perform a rotation at 1 and fix the colors so that 2 is a valid 4-node. Because 2 and 3 are oriented the same way, a single rotation will fix the problem.


    The next case is exactly the same except for the color of 0. If 0 is red then all that needs to be done is a simple color flip to restore the red black condition. The color flip is made by changing 0 and 2 to black and 1 to red:
    Code:
    Case 2:
    
       1B                 1R
    
    0R    2R     --->  0B    2B
    
             3R                 3R
    The next case is similar to case 1 except with 2 and 3 oriented in different directions. This case requires a rotation to turn the node into a valid 4-node. If you are yelling "double rotation!" you are correct. :-)
    Code:
    Case 3:
    
        1B                1B                       3B
    
    0B      2R  --->  0B      3R     --->      1R      2R
    
         3R                      2R         0B
    The final case is identical to case 3 except 0 is red. Just like case 2, all that is needed to restore the red black condition is a simple color flip:
    Code:
    Case 4:
    
        1B                   1R
    
    0R      2R     --->  0B      2B
    
         3R                   3R
    The reason we did not need a status flag to tell us when to rebalance is because in this implementation, the rebalancing is done at every step. However, only when the conditions are met for a rule 2 violation will anything actually be done. The result is a slightly ugly collection of conditionals:
    Code:
    struct node *rebalance_insert ( struct node *tree, int dir )
    {
      struct node *child = tree->link[dir];
      struct node *sibling = tree->link[!dir];
    
      /* Same direction case */
      if ( child->link[dir] != NULL ) {
        if ( child->red && child->link[dir]->red ) {
          if ( is_black ( sibling ) ) {
            tree = single_rotate ( tree, dir );
    
            tree->red = 0;
            tree->link[0]->red = tree->link[1]->red = 1;
          }
          else {
            tree->red = 1;
            tree->link[0]->red = tree->link[1]->red = 0;
          }
        }
      }
    
      /* Opposing direction case */
      if ( child->link[!dir] != NULL ) {
        if ( child->red && child->link[!dir]->red ) {
          if ( is_black ( sibling ) ) {
            tree = double_rotate ( tree, dir );
    
            tree->red = 0;
            tree->link[0]->red = tree->link[1]->red = 1;
          }
          else {
            tree->red = 1;
            tree->link[0]->red = tree->link[1]->red = 0;
          }
        }
      }
    
      return tree;
    }
    This code is a direct translation of the diagrams into C. First it checks for the two cases in the same direction, then the two cases in the opposing direction. Thanks to the use of dir, we can avoid the symmetric cases that cause balanced binary search tree routines to be so bloated and intimidating. rebalance_insert is very simple and should be easy to follow regardless of experience. If you have gotten this far in the article, you should not have any problems with it. :-)


    Exercises

    1) Rewrite rebalance_insert to avoid relying on conditionals. Note: This may require sweeping changes to insert_node_r.

    2) Rewrite insert_node without recursion.

    3) Verify that insertion into a Red Black tree never requires more than two rotations.



    Deletion

    Just as removing a node from an AVL tree was more complicated than insertion, so it is with Red Black trees. We will use a similar recursive removal algorithm as the one for AVL trees, with the added necessity of an auxiliary function for maintaining rule 3. On top of that, we will also need a status flag to tell the algorithm when to rebalance. We were saved from that with insertion, but with deletion it is much easier this way. The code from remove_node is simple:
    Code:
    struct node *remove_node ( struct node *tree, int data )
    {
      int new_height;
    
      tree = remove_node_r ( tree, data, &new_height );
    
      if ( tree != NULL && tree->red )
        tree->red = 0;
    
      return tree;
    }
    It should be familiar, but note that the test to see if the root is red must now include a test for a null pointer. There is the possibility that remove_node might be removing the root, and that the root is the only item in the tree. In which case, the return from remove_node_r will be an empty tree, or a null pointer.

    remove_node_r uses the same recursive deletion-by-copying algorithm that we used for AVL deletion. The only difference is that now we need to consider two cases when deleting the inorder predecessor. If the inorder predecessor is red then we can safely delete it without having to rebalance the tree. The reason for this is because removing a red node cannot violate either rule 1 or rule 2, so there is no reason to set the flag and call rebalance_remove. On the other hand, if the inorder predecessor is black then that will violate rule 1 by changing the black height of that path. In that case we need to set the status flag.

    But there's more! If the node that replaces the black node that we just removed is red, we can simply recolor it black and clear the status flag. Rebalancing accomplished without calling rebalance_remove! Here is the code:
    Code:
    struct node *remove_node_r ( struct node *tree, int data, int *new_height )
    {
      if ( tree == NULL )
        *new_height = 0;
      else if ( data == tree->data ) {
        if ( tree->link[0] != NULL && tree->link[1] != NULL ) {
          struct node *heir = tree->link[0];
    
          while ( heir->link[1] != NULL )
            heir = heir->link[1];
    
          tree->data = heir->data;
    
          tree->link[0] = remove_node_r ( tree->link[0], tree->data, new_height );
    
          if ( *new_height )
            tree = rebalance_remove ( tree, 0, new_height );
        }
        else {
          struct node *save = tree;
    
          *new_height = !tree->red;
    
          tree = tree->link[tree->link[0] == NULL];
          destroy_node ( save );
    
          if ( tree != NULL && tree->red ) {
            tree->red = 0;
            *new_height = 0;
          }
        }
      }
      else {
        int dir = data > tree->data;
    
        tree->link[dir] = remove_node_r ( tree->link[dir], data, new_height );
    
        if ( *new_height )
          tree = rebalance_remove ( tree, dir, new_height );
      }
    
      return tree;
    }
    Notice the convoluted tests when deleting the inorder predecessor:

    1) Save the soon to be deleted node.
    2) Set or clear new_height based on the color of the doomed node.
    3) Execute the node.
    4) If the replacement is red, make it black and clear new_height.

    Other than that, the algorithm is identical to the deletion routine we wrote for AVL trees. The fun part is in rebalance_remove, where rotations are made and colors are flipped.

    The rebalancing of a Red Black deletion consists of four cases. The first case is if the other child of the node (tree->link[!dir] rather than tree->link[dir]) is red, we enable the other cases with a single rotation. As it turns out, rebalancing is much easier if the sibling is black.
    Code:
    Case 1:
    
        1B                      3B
    
    0B      3R     --->     1R      4B
    
         2B    4B        0B    2B
    It is important to remember that this case propagates instead of clearing the status flag. This step is only a convenience step to make the real rebalancing easier. The next case actually does something; it propagates the need for rebalancing up the tree without a rotation. How? Remember that the violation we are dealing with is that of black height. If the sibling is black (as we guaranteed previously) and both of its children are black then we can recolor the sibling red and move up.
    Code:
    Case 2 (A color of E means either red or black):
    
        1E                   1E
    
    0B      3B     --->  0B      3R
    
         2B    4B             2B    4B
    The next case is a two part case. If the opposing child of 3 is red then we single rotate at 3. This is the first half of a double rotation. If the opposing child of 3 is not red then we single rotate at 1. This is the second half of a double rotation, or of the first half was not needed, it is a stand-alone single rotation.
    Code:
    Case 3a:
    
        1E                1E
    
    0B      3B  --->  0B      2R
    
         2R                       3B
    Code:
    case 3b:
    
        1E                     2E
    
    0B      2R     -->     1B      3B
    
               3B       0B
    Here is the code that performs this magic:
    Code:
    struct node *rebalance_remove ( struct node *tree, int dir, int *new_height )
    {
      struct node *sibling = tree->link[!dir];
    
      /* Make sibling black case */
      if ( sibling->red ) {
        sibling->red = 0;
        tree->red = 1;
    
        tree = single_rotate ( tree, !dir );
    
        tree->link[dir]->red = 0;
        tree->link[dir]->link[!dir]->red = 1;
      }
      else {
        /* Zero rotation case */
        if ( is_black ( sibling->link[0] ) && is_black ( sibling->link[1] ) )
          sibling->red = 1;
        else {
          /* Double rotation case */
          if ( is_black ( sibling->link[!dir] ) ) {
            sibling->link[dir]->red = 0;
            sibling->red = 1;
    
            tree->link[!dir] = single_rotate ( tree->link[!dir], dir );
    
            sibling = tree->link[!dir];
          }
    
          /* Single rotation case */
          sibling->red = tree->red;
          tree->red = 0;
          sibling->link[!dir]->red = 0;
    
          tree = single_rotate ( tree, !dir );
    
          *new_height = 0;
        }
      }
    
      return tree;
    }
    I strongly recommend that you study the code and try to understand what it does and why. Red Black trees are more obscure than AVL trees and misconceptions can easily be made. I suggest working out several examples on paper while following the code. Make sure that the algorithm does what you expect, otherwise try to figure out why.


    Exercises

    1) Rewrite rebalance_remove to separate the single and double rotation cases. Is this an improvement?

    2) Rewrite remove_node without using recursion.

    3) Verify that deletion from a Red Black tree never requires more than three rotations.



    Splaying

    Splaying is the technique of moving a node to the root of a binary search tree by two levels at a time. Remember that root insertion from Part I did something similar except with single rotations at each level. Splaying involves single and double rotations, but the result is a drastically improved structure because while root insertion brings a node to the root, it also pushes another node along the path just as far down. With splaying, the node is lifted to the root and the height of every node along the search path is roughly halved. There are three cases for splaying (ignoring symmetric cases):
    Code:
    Zig
    
    A            A
         --->
      B        B
    Code:
    Zig-Zig
    
    A               B               C
    
      B    --->  A     C  --->    B
    
        C                       A
    Code:
    Zig-Zag
    
    A          A               C
    
      B  --->    C    --->  A     B
    
    C              B
    Zig and Zig-Zag are identical to the single and double rotations we have been working with, respectively. Zig-Zig is different in that the first rotation is at the grandparent of the node rather than the parent.

    By splaying for every search in a binary search tree, we can guarantee amortized logarithmic performance. In fact, splay trees are often comparable in performance with AVL and Red Black trees. Once again, that is amazing considering the fact that splay trees maintain no balance factors.

    The analysis of a splay tree is far beyond me, so we'll just move on to a description of the two ways to splay, bottom up and top down.



    Bottom Up Splaying

    Bottom up splaying is very simple. Walk down the tree until you find the node that you have been looking for and then splay it back up. There are several boundary conditions that should be kept in mind though.

    The first boundary condition is that of splaying for a nonexistent item. As we have been using, a status flag is ideal for this task. If the search path gets to a null pointer, clear the flag, otherwise the item was found and we set the flag. This is contrary to the original design of splaying, where an unsuccessful search would result in splaying the leaf on the search path that terminated the search, but it can be useful at times. We will study both variations.

    The second boundary condition is walking back up we need to be at least the node's grandparent for splaying to work properly. Otherwise null pointers will be dereferenced and all manner of ghoulies show up. The solution to this is the same as we used for Red Black insertion, just make sure that there are three nodes to splay using conditionals. :-)

    The final case is when the last splay does not bring the item to the root of the tree. This could happen if there are not three nodes to splay near the root. When this is the case, the only possible reason is because the item is a child of the root or it is nonexistent. In the case where the item is a child of the root, a single zig will suffice. In the case where the item is nonexistent, we need to decide whether to splay or not.

    Assuming we have a recursive splay_r function, we can write a splay function that does not splay on search misses immediately:
    Code:
    struct node *splay ( struct node *tree, int data )
    {
      int splay = 0;
    
      tree = splay_r ( tree, data, &splay );
    
      /* Final Zig if necessary */
      if ( splay && data != tree->data ) {
        int dir = data > tree->data;
    
        tree = single_rotate ( tree, dir );
      }
    
      return tree;
    }
    The splay flag will be true if the item was found, and if it isn't the root node then a zig is required. Otherwise everything supposedly worked and the root is the item we were searching for. Notice how we reused single_rotate again. Aren't you glad it wasn't written to modify AVL balance factors too? :-)

    In the case where we want to be strictly correct according to the original paper on splaying, we can rewrite splay to handle search misses by splaying the last valid node on the search path:
    Code:
    struct node *splay ( struct node *tree, int data )
    {
      int dir;
    
      tree = splay_r ( tree, data );
    
      dir = data > tree->data;
    
      /* Final Zig if necessary */
      if ( tree->link[dir] != NULL && data != tree->data )
        tree = single_rotate ( tree, dir );
    
      return tree;
    }
    Without the splay flag, we need to make sure that a rotation is possible without dereferencing null pointers. We could to this in the single_rotate function, but that could potentially hide other bugs because single_rotate is so fundamental of an operation with balanced binary search trees. So we add a single test to see if the root has a child. If it does then we can rotate and be confident that the new root is either the item we wanted or the previous item along an unsuccessful search path. If the root does not have a child then there is no need to rotate because the item we want is at the root.

    The splay_r function is just a simple binary search tree search. The only difference is that on the return trip, we splay the node if it was found for the first variation of splaying:
    Code:
    struct node *splay_r ( struct node *tree, int data, int *splay )
    {
      if ( tree == NULL )
        *splay = 0;
      else if ( data == tree->data )
        *splay = 1;
      else {
        int dir = data > tree->data;
    
        tree->link[dir] = splay_r ( tree->link[dir], data, splay );
    
        if ( *splay ) {
          if ( tree->link[dir] != NULL ) {
            struct node *zig_target = tree->link[dir]->link[dir];
            struct node *zag_target = tree->link[dir]->link[!dir];
    
            if ( zig_target != NULL && data == zig_target->data ) {
              /* Zig-Zig */
              tree = single_rotate ( tree, dir );
              tree = single_rotate ( tree, dir );
            }
            else if ( zag_target != NULL && data == zag_target->data ) {
              /* Zig-Zag */
              tree->link[dir] = single_rotate ( tree->link[dir], !dir );
              tree = single_rotate ( tree, dir );
            }
          }
        }
      }
    
      return tree;
    }
    The code for splaying on an unsuccessful search is even cleaner!
    Code:
    struct node *splay_r ( struct node *tree, int data )
    {
      if ( tree == NULL )
        ;
      else {
        int dir = data > tree->data;
    
        tree->link[dir] = splay_r ( tree->link[dir], data );
    
        if ( tree->link[dir] != NULL ) {
          struct node *zig_target = tree->link[dir]->link[dir];
          struct node *zag_target = tree->link[dir]->link[!dir];
    
          if ( zig_target != NULL && data == zig_target->data ) {
            /* Zig-Zig */
            tree = single_rotate ( tree, dir );
            tree = single_rotate ( tree, dir );
          }
          else if ( zag_target != NULL && data == zag_target->data ) {
            /* Zig-Zag */
            tree->link[dir] = single_rotate ( tree->link[dir], !dir );
            tree = single_rotate ( tree, dir );
          }
        }
      }
    
      return tree;
    }
    Notice that I continue to use the same if test even if there is nothing to do when tree is a null pointer. This is simply to keep the code consistent and easier to follow for the purposes of this article. Also note that thanks to dir, we can avoid those nasty symmetric cases. While Zig is supposed to be left and Zag is supposed to be right, I use them as dir and !dir to save commenting time, if you have not already figured that out.

    Nothing in this code should be problematic for you by now, so please study it and do a few test runs. Bottom up splaying is easy, but top down splaying, which we will study next, can be confusing.

    Exercises

    1) Can you think of any way to remove some of the special cases that we had to handle?

    2) Rewrite splay using a bottom up algorithm without using recursion.



    Top Down Splaying

    Top down splaying is best done iteratively. Bottom up splaying requires a downward search of the tree followed by an upward march while splaying. The problem is that we need to handle several special cases. Top down splaying removes those special cases at the cost of clarity.

    To perform top down splaying, we walk down the tree, and for every node we use a second tree to hold temporary links. The left temporary holds nodes that are less than the root but not in either of its subtrees and the right temporary holds nodes that are greater than the root but not in either of its subtrees. For each case where there are two nodes less than or greater than the root, a rotation is performed. The operations are as follows.
    Code:
    Zig
    =====================================
    Before Zig:
    
    left            root            right
      ~              A                ~
    
                  a     B
    
                      b   c
    
    
    After Zig:
    
    left            root            right
      ~              B                A
    
                   b   c            a
    At every step along the search path, root is the root of the subtree being worked on. Left and right are the trees that hold the nodes less than and greater than root, but not in its subtrees, respectively. If the operation is a Zig, then the current root (A) is stored in either left or right and the child of the current root (B) becomes the new root for splaying.
    Code:
    Zig-Zig
    =====================================
    Before Zig-Zig:
    
    left            root            right
      ~              C                ~
    
                  B     d
    
               A    c
    
             a   b
    
    After Zig-Zig:
    
    left            root            right
      ~              A                B
    
                   a   b                 C
    
                                       c   d
    If the operation is a Zig-Zig, then a single rotation is performed on the current root (C), and the root of the resulting tree (B) is stored in either left or right. The child of the new root (A) becomes the new root for splaying.
    Code:
    Zig-Zag
    =====================================
    Before Zig-Zag:
    
    left            root            right
      ~              C                ~
    
                 A       d
    
              a     B
    
                  b   c
    
    After Zig-Zag:
    
    left            root            right
      A              B                C
    
    a              b   c                d
    If the operation is a Zig-Zag, then the current root (C) is stored in either left or right, its child (A) is stored in the other temporary tree, and its child (B) becomes the new root for splaying.

    When splaying is complete, the current root’s children are added to the left and right temporary trees. Then the left temporary tree becomes the current root’s left subtree and the right temporary tree becomes the current root’s right subtree.

    At each step, we need to consider the case where the current root has no children. First, when considering a Zig, if the root has no child along the path then splaying must terminate. Second, after a single rotation, if the root has no child along the same path, splaying must again terminate. Both of these cases are meant to avoid dereferencing invalid links and are also used as convenient times to reassemble the tree and return.
    Code:
    struct node *splay ( struct node *tree, int data )
    {
      struct node N = { 0, { 0, 0 } };
      struct node *l = &N, *r = &N;
      struct node **save;
    
      if ( tree == NULL )
        return tree;
    
      while ( data != tree->data ) {
        int dir = data > tree->data;
    
        if ( tree->link[dir] == NULL )
          break;
    
        if ( ( dir == 0 && data < tree->link[dir]->data )
          || ( dir != 0 && data > tree->link[dir]->data ) )
        {
          tree = single_rotate ( tree, dir );
    
          if ( tree->link[dir] == NULL )
            break;
        }
    
        save = dir ? &l : &r;
        (*save)->link[dir] = tree;
        *save = tree;
        tree = tree->link[dir];
      }
    
      /* Reassemble */
      l->link[1] = tree->link[0];
      r->link[0] = tree->link[1];
      tree->link[0] = N.link[1];
      tree->link[1] = N.link[0];
    
      return tree;
    }
    Notice that I used the save variable to pick the left or right temporary trees. This is another attempt at avoiding symmetric cases, but it shouldn't cause any trouble as we covered this technique in Part I.


    Exercises

    1) Rewrite splay with all of the symmetric cases. Is the author crazy for writing the code without them? :-)

    2) Execute the code with several example trees. Does it do what you expected?

    3) Rewrite splay with a top down algorithm using recursion.



    Splay Trees


    Insertion

    Insertion into a splay tree is dreadfully simple if you already have the splaying function. The code is so simple I'll leave figuring it out as an exercise:
    Code:
    struct node *insert_node ( struct node *tree, int data )
    {
      struct node *walk = make_node ( data );
    
      if ( walk == NULL || tree == NULL )
        return walk;
    
      tree = splay ( tree, data );
    
      if ( data == tree->data ) {
        destroy_node ( walk );
    
        return tree;
      }
      else {
        int dir = data > tree->data;
    
        walk->link[dir] = tree->link[dir];
        walk->link[!dir] = tree;
        tree->link[dir] = NULL;
    
        return walk;
      }
    }

    Deletion

    Deletion from a splay tree is even simpler than insertion (for once!). Once again, the code is so simple that figuring it out is left as an exercise:
    Code:
    struct node *remove_node ( struct node *tree, int data )
    {
      struct node *save;
    
      if ( tree == NULL )
        return NULL;
    
      tree = splay ( tree, data );
    
      if ( data == tree->data ) {
        if ( tree->link[0] == NULL )
          save = tree->link[1];
        else {
          save = splay ( tree->link[0], data );
          save->link[1] = tree->link[1];
        }
    
        destroy_node ( tree );
        tree = save;
      }
    
      return tree;
    }

    Conclusion

    In this article we looked at several types of balanced binary search tree that guarantee logarithmic performance. You learned that the algorithms are not nearly as complicated as others may have told you, and you also saw how to implement insertion and deletion with all of them.

    Balanced binary search trees are very useful and I hope that this article was helpful in alleviating some of the mysteries. :-)
    My best code is written with the delete key.

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