Thread: memmory addresses not working out or me screwing somthing up

my plan is to start of with three nodes to the right. set some variable to 1st 2nd and 3rd nodes and see if i can make it do what i want. then the next stage is to add a node to the root nodes left and try rotate to the left. each time adding nodes till i end up with 5 or 6 nodes with the node being moved to the left is somewhere in the middle. then once i have mastered that do it all with right rotation. then see if i can call the same function twice for double left... if not sit and work out that from the beginning. to do all this weather the tree is balanced or not is unimportant at this stage its just about getting the rotation functions correct

then once all that is working i can start work on the actual balancing factors and only have to plug in the appropriate function

@cooper1200,

There is no spoon (from "The Matrix").

In other words, there is no tr. I know, you have one, it's Bin_Tree, but look at your insert_node. It takes a Bin_Node **. Inside insert_node, there is no tr, no Bin_Tree.

You don't need it.

Bin_Tree has a root, I know.

When you call insert_node, you call that from insert (where you do have Bin_Tree). But look at that call.

Code:

 insert_node( &root, node );

Actually, that's a mistake. It shouldn't be the temporary Bin_Node *root;

That call should be

Code:

insert_node( &( tree->root ), node );

I had not noticed that before. Actually, I recall mentioning it, but never mind...it should be changed to the tree->root, not root.

Now, look at the insert_node signature:

Code:

 bool insert_node( Bin_Node ** root, Bin_Node *node )....

root is a pointer to a pointer, which is modified with something like

*root = node;

Or whatever context is appropriate. That will assign Bin_Tree's node. Insert_Node has (or will have when you fix that as I pointed out) Bin_Tree's root.

Via pointer to a pointer (the **)

im not getting this. why is i changed it was set to p

tree is empty
i send in node 1
it returns

were not done yet as i kept finding out
i then have to set tree->root to node 1

i send in node 2
it finds node one and goes to node 1's right and sends in node 1's address
inserts 2 by *root = node
returns
i send node three in
it finds node 1 and goes to node 1's right and sends in node 1's address
it finds node 2 and goes to node 2's right and sends in node 2's address
inserts 3 by *root = node
returns

each time i have sent in the address of the parent node.... r more precisely i have sent in the address of where the parent nodes right pointer is

like you did with your function you sent in a pointer to a pointer and then changed the value it pointed at... ie in the case of the tree it was pointing at null but once we get to the point of insertion it gets set to point to the node

If I follow you, yes, that's it.

were typing at the same time here lol

so following that train of thought to set the tree root to point to two i have to know what was pointing at 1

In yet another way to say it, &(tree->root) looks like a parent inside insert_node, and &((*root)->left) makes that Bin_Node's left look like a parent inside insert_node

oh wait i think the penny has dropped (yes there was a loud echoing clang) i am rotating node to the left.... so **root is whats pointing at node
in other words 1 is node node->left is 2 and node->left->left is 3
so to make what ever is pointing at node 1 i would make *root point at 2

you beat me to it lol

ok to take your examples further....

Code:

f1( **p) { f2(&(*p))??
f2 (**p) or (***p)

*** is gettin' a little nuts


I'm going to have to think about f2 or ***p over another cup of coffee....the earlier ones are all used up