Binary Tree Revisited: Near Completion

Nakeerb · 01-21-2003, 12:23 AM

With the help of Nick I am coming very close to being able to insert data into this binary tree... yet some parts still refuse to work.

Here is the entire code segment thus far

Code:

#include <iostream>
#include <ctype>
#include <conio>
using namespace std;


struct Token {
    char type;
    double val;
};


struct node{
	node();
	char data;
	node *left, *right;
};


node::node(){
	left = NULL;
	right = NULL;
}


void read_next_token(Token &next_token);
node* start(Token &next_token);
node* expr(Token &next_token);
node* term(Token &next_token);
node* factor(Token &next_token);
void traverseInOrder(node *root);


int main(){
    node* val;
	Token next_token;
	
	cout << "Parenthesized Infix equation: ";
	
    read_next_token(next_token);   
    val = start(next_token);       
    cout << "Result = ";
	traverseInOrder(val);
	getch();
    return 0;
}  //end main
        

void read_next_token(Token &next_token){
    double c;

    c = cin.get();    //doesn't require an enter press

    if (isdigit(c)) {
        next_token.type = '#';         
        next_token.val  = c - 48;    //to get rid of the ascii value offset
        return;
    }
	
	next_token.type = char(c);   
}  


node* start(Token &next_token){
    node* val = expr(next_token);  
    return val;                     
}


node* expr(Token &next_token){
    node* val = new node;

    if (val -> left == NULL)
   				val -> left = term(next_token);
			else
				val -> right = term(next_token); 
	   
	
    switch(next_token.type){
	
	    case '+':
		case '-':
	        read_next_token(next_token);
	        if (val -> left == NULL)
   				val -> left = expr(next_token); 
			else
				val -> right = expr(next_token); 
	        break;
	     
			
	 
			
	    default:
	        break;
    }
 
    return val;
}


node* term(Token &next_token){
    node* val = new node;
	//zerotest;

	if (val -> left == NULL)
    val -> left = factor(next_token);
	else
	val -> right = factor(next_token); 
	
    switch(next_token.type){
	
	    case '*':
		case '/':
	        read_next_token(next_token);  //get the next inputted letter
			
	        if (val -> left == NULL)
   				val -> left = term(next_token); 
			else
				val -> right = term(next_token); 
	        break;
			
	
			
	    default:
	        break;
    }

    return val;
}
    

node* factor(Token &next_token){
    node* val = new node;
    
    switch(next_token.type){
	
	    case '#':                         //if the type of the token is a number
	        val -> data = char(next_token.val);       
	        read_next_token(next_token);  //get the next inputted letter
	        break;
			
	    case '(':            //the only other thing it can be here is a (
	        read_next_token(next_token);  //get the next inputted letter
	        if (val -> left == NULL)
   				val -> left = expr(next_token); 
			else
				val -> right = expr(next_token); 
	        read_next_token(next_token);  //get the next inputted letter
	        break;
			
	    default:
	        break;
    }
    
    return val;  
}


void traverseInOrder(node *root){
	
	if (!root)		//test first to see if done
		return;
		
	traverseInOrder(root->left);
	cout << root->data << " ";	
	traverseInOrder(root->right);	   
}

Nick: Now I understand how this "descended recurse" works. Very spiffy :P Unfortunately I am now having trouble converting this to the node insertions. Anyone able to tell me why I am not able to view the data I desire when I do an inOrder traversal? Perhaps I am not inserting correctly...

Nakeerb · 01-21-2003, 12:28 AM

Alright there we go. Alignments fixed

Nakeerb · 01-21-2003, 01:27 AM

Does this have anything to do with checking for left?

Nick · 01-21-2003, 01:44 AM

In each function you just look at the different cases
that can arise and return a tree representing them.
You only need to use the constructor
node(char data, node* left, node* right). This way
you return a tree even if it has incorrect data.

In most of the functions all you need is
node* root, *right, *left. So for example if your
parsing 5 + 6 in expr you would create a tree for it by
doing
left = term();
right = expr();
root = new node('+', left, right);

Be aware that if your just parsing
4;
your function must basically do
left = term();
root = left;
This will go through the default case in expr and term.

vasanth · 01-21-2003, 06:31 AM

i dint go through your entire program.. but i feel your traversal is wrong... You have to start from the root but display it only after the leaf node is done etc.. fro different taversal...

example

Code:

FOR PRE ORDER DISPLAY

tree::predisplay(node *te)
{

cout<<"\n"<<te->key;

if(te->left!=NULL)
predisplay(te->left);

if(te->right!=NULL)
predisplay(te->right);

return SUCCESS;
}


POST ORDER DISPLAY

tree::postdisplay(node *te)
{

if(te->left!=NULL)
postdisplay(te->left);

if(te->right!=NULL)
postdisplay(te->right);

cout<<"\n"<<te->key;

return SUCCESS;
}


INORDER DISPLAY
tree::indisplay(node *te)
{

if(te->left!=NULL)
indisplay(te->left);

cout<<"\n"<<te->key;

if(te->right!=NULL)
indisplay(te->right);



return SUCCESS;
}

so all these functions are called with the root as argument.. and what is !root supposed to do...

Nick · 01-21-2003, 02:38 PM

His traversal is correct. As long as the tree is formed
correctly it should run. But I think passing next_token
into each function is ugly. Better to use a class or use
a global variable.

You should probably skip white space at the beginning of
read_next_token and I'm not sure why you are using double c. Try to test each function in isolation. Simply test read_next_token then test factor, term, expr in that order.

Bakamonoda · 01-21-2003, 03:12 PM

Nakeerb here, temporary name.

Nick, then is all the new code I wrote a waste? I understand how you mean but I have a feeling I'd have to clean up SO much code to get it working. The ( and ) are not supposed to be part of the binary tree at all. only numbers and operators, and all evaluations are FULLY parenthesized.

As in you'll never see (3 + 2 + (4 + 5)), it'd be ((3 + 2)+(4+5))

However, I have too much crappy code written in alerady. What would I do in place of something such as this:

Code:

 switch(next_token.type){
	
	    case '#':                         //if the type of the token is a number
	        val -> data = char(next_token.val);       
	        read_next_token(next_token);  //get the next inputted letter
	        break;
			
	    case '(':            //the only other thing it can be here is a (
	        read_next_token(next_token);  //get the next inputted letter
	        if (val -> left == NULL) //EQUAL TERMS YOU FOOL
   				val -> left = expr(next_token); 
			else
				val -> right = expr(next_token); 
	        read_next_token(next_token);  //get the next inputted letter
	        break;
			
	    default:
	        break;
    }

If you can help me with this function I will do the rest. I am just totally lost. Do I need to declare two node pointers *val and *root? Or declare root when I... auugh! I couldn't sleep last night because of this stupid binary tree problem O.o;;

when you say left, I assumed you meant a temp node's left is equal. Then you pass the temp node's left into the constructor... oyyy, no idea.

Nick · 01-21-2003, 03:48 PM

Code:

switch(next_token.type){
    case '#':                         //if the type of the token is a number
        val -> data = char(next_token.val);       
        read_next_token(next_token);  //get the next inputted letter
        break;
    case '(':            //the only other thing it can be here is a (
        read_next_token(next_token);  //get the next inputted letter
        if (val -> left == NULL) //EQUAL TERMS YOU FOOL
	val -> left = expr(next_token); 
        else
	val -> right = expr(next_token); 
	 read_next_token(next_token);  
                 break;
        default:
                 break;
}

You could do something like this

Code:

node* factor(Token &next_token){
    node* root = NULL;
     
    switch(next_token.type){
    case '#':                         
           root = new node(next_token.val, NULL, NULL);       
           read_next_token(next_token);  
           break;
     case '(':           
           read_next_token(next_token); 
           root = expr(next_token); 
           read_next_Token(next_token);
     default:
           break;
    }
    
    return root;  
}

Nakeerb · 01-21-2003, 06:19 PM

So I can do this for all functions? When would I need to append to right side or left?

Nick · 01-21-2003, 08:10 PM

This is just for factor. The functions like expr there
are cases where you would set root = term() (this is
in the default cases) and there
are cases where you would do something like
node* left = term();
node* right = expr();
root = node('+', left, right)

Nakeerb · 01-21-2003, 08:35 PM

I rewrote another program that does the same based on the psuedocode I was given today at around noon. However it still does not work. I know I am going off on an extreme tangent by posting this, but any help would be appreciated.

Code:

#include <iostream>
#include <conio>
#include <ctype>
#include <cstdlib>
using namespace std;

class BinaryTree{
	public:
		
		~BinaryTree(){
		
			if (left != NULL)
				delete left;
				
			if (left != NULL)
				delete right;
				
		}
		
	BinaryTree *left;     //pointer to a left subtree
	BinaryTree *right;    //pointer to a right subtree
	char data;            //node data
};

BinaryTree* buildTree(char *infix, int &pos);
void preOrderTraverse(BinaryTree *root);
void postOrderTraverse(BinaryTree *root);
double evaluate(BinaryTree *root);
bool isOperator(char ch);
void changecolor(char* word, int color);
void flushin();


int main(){
	char infix[50], yesOrNo;
	BinaryTree *tree;
	int pos = 1;
	
	do{
		clrscr();
		changecolor("Binary Expression Tree\r\n\r\n", LIGHTGREEN);
		cout << "Give me a <fully> parenthesized infix equation: \n";
		changecolor("\r\nInfix Expression: ", WHITE); 
		cin.getline(infix, '\n');
		cout << endl << endl;
		
		tree = buildTree(infix, pos);      //inserts infix expression into a binary tree
		cout << "Prefix form of infix: "; 
		preOrderTraverse(tree);            //outputs prefix form of infix
		cout << "\nPostfix form of infix: "; 
		postOrderTraverse(tree);           //outputs postfix form of infix
		 
		cout << "\n\nEvaluation of tree: ";
		evaluate(tree);
		
		delete tree;   //returns the memory to the heap
		
		cout << "\nAgain? ";
		
		do{
			cin.get(yesOrNo);
		}while((tolower(yesOrNo) != 'y') && (tolower(yesOrNo) != 'n'));
		
		flushin();   //clear input buffer
	}while(tolower(yesOrNo) == 'y');
	
	return 0;
}


BinaryTree* buildTree(char *infix, int &pos){
	BinaryTree *root = new BinaryTree;
	
	if(infix[pos] == '(')
		root -> left = buildTree(infix, ++pos);    //recursive call: builds a subtree to the left
	
	else{
		root -> left = new BinaryTree;
		cout << infix[pos] << " being added to left branch...\n";
		root -> left -> data = infix[pos];         //assigns data unit
	}
	
	pos++;                        //so we are able to get the next character of infix
	root -> data = infix[pos];    //assigns the operator to the root's data character
	pos++;                        //so we are able to get the next character of infix
	
	if (infix[pos] == '(')
		root -> right = buildTree(infix, ++pos);   //recursive call: builds a subtree to the right
	
	else{
		root -> right = new BinaryTree;
		cout << infix[pos] << " being added to right branch...\n";
		root -> right -> data = infix[pos];
	}
	
	pos++;       //skip over the ending ) bracket
	return root;
		
}


void preOrderTraverse(BinaryTree *root){
	cout << (root -> data);			   //output the data of the root
	
	if (root -> left)
		preOrderTraverse(root -> left);  //traverse down to the left side of the tree
		
	if (root -> right)	  
		preOrderTraverse(root -> right); //traverse down to the right side of the tree	
}


void postOrderTraverse(BinaryTree *root){	 

	if (root -> left)
		postOrderTraverse(root -> left);  //traverse down to the left side of the tree
	
	if (root -> right)	 	 
		postOrderTraverse(root -> right); //traverse down to the right side of the tree
		
	cout << (root -> data);	      //output the data of the root
}


double evaluate(BinaryTree *root){
	double temp, divisor;
	
	if (isdigit(root -> data))       //if the data character is a number
		temp = (root -> data) - 48;  //convert it's ascii value to a format a double can recognize
		
		
	if (isOperator(root -> data)){   //if the data character is an operator
	
		switch (root -> data){   //apply the operator to the left and right values using inorder recursive traversal
			case '+': temp = evaluate(root -> left) + evaluate(root -> right); break;
			case '-': temp = evaluate(root -> left) - evaluate(root -> right); break;
			case '*': temp = evaluate(root -> left) * evaluate(root -> right); break;
			case '/':
				divisor = evaluate(root -> right);
				
				if (divisor == 0){   //cannot divide by 0
					cout << "\nDivide by 0 error.\n";
					exit(1);
				}
				
				else
					temp = evaluate(root -> left) / divisor;
					
				break;
				
			case '^':	 temp = pow(evaluate(root -> left), evaluate(root -> right)); break;
			default: cout << "\nError, operator unknown\n"; break;	
			  
		}
	
	}	 
		
	return temp;
}


bool isOperator(char ch){
	string acceptedOperators = "+-*/^";
	
	if(acceptedOperators.find(ch, 0) != string::npos)    //if ch can't be found in the string
		return true;
		
	else
		return false;
		
} //end bool isOperator


void changecolor(char* word, int color){
	textcolor(color); 		 //changes text to the color given through the argument
	cprintf("%s", word);	   	    //outputs the string in the given color
	//cprintf(word);
	textcolor(LIGHTGRAY);	 //change color back to the default light gray
} //end void changecolor


void flushin(){

	while (cin.get() != '\n'){}
	
} //end void flushin

Nick · 01-21-2003, 09:54 PM

It took me only about 10 minutes
to add the few lines to build the tree (:

These are the two most important functions to build the
tree the rest of the functions follow the same form.

Code:

Node* term()
{
    Node* root, *left, *right;
    
    left = factor();

    switch(next_token.type) {
    case MULTIPLY:
        match(MULTIPLY);
        right = term();
        root = new Node('*', left, right);
        break;
    case DIVIDE:
        match(DIVIDE);
        right = term();
        root = new Node('/', left, right);
        break;
    default:
        root = left;
        break;
    }

    return root;
}
    

Node* factor()
{
    Node* root = 0;
    
    switch(next_token.type) {
    case NUMBER:
        root = new Node('0' + next_token.val, 0, 0);
        match(NUMBER);
        break;
    case LEFT_PARAN:
        match(LEFT_PARAN);
        root = expr();
        match(RIGHT_PARAN);
        break;
    default:
        break;
    }
    
    return root;
}

Nakeerb · 01-21-2003, 11:32 PM

Code:

//This is the header file "binary_tree.h"


//===========================/=/=/=/=/=/=/=========================//
//                            STRUCTURES                           //
//===========================/=/=/=/=/=/=/=========================//

#ifndef BINARY_TREE_H
#define BINARY_TREE_H


struct BinaryTree{    //contains the neccessary functions and data for a functioning binary expression tree

	BinaryTree();   //Default constructor
	//Precondition: Object has been declared with no arguments
	//Postcondition: The left and right pointers have been set to NULL
	
	~BinaryTree();  //Destructor
	//Precondition: The object exists, and has been deleted by "delete"
	//Postcondition: The left and right pointers' memory are sent back to the heap
	
	char data;            //node data (can be either a digit or an operator)
	BinaryTree *left;     //pointer to a left subtree
	BinaryTree *right;    //pointer to a right subtree
};  //end struct BinaryTree


#endif

Code:

//===========================/=/=/=/=/=/=/=========================//
//                              INCLUDES                           //
//===========================/=/=/=/=/=/=/=========================//


#include <iostream>       //for the insertion operators, cout, and cin
#include <conio>          //for the color functions and clrscr()
#include <ctype>          //for the isdigit() and tolower() functions
#include <cstdlib>        //for the exit() function 
#include <cmath>          //for the pow function
#include "binary_tree.h"  //contains the struct for the binary tree
using namespace std;


//===========================/=/=/=/=/=/=/=========================//
//                        FUNCTION PROTOTYPES                      //
//===========================/=/=/=/=/=/=/=========================//


BinaryTree* buildTree(char *infix, int &pos);
//Precondition: infix has been given an infix equation string by the user, and
//pos has been given a value (to begin with, it is 1)
//Postcondition: A binary expression tree is created, returning the root from 
//a fully parenthesized expression given through infix. It stores digits under
//left and right nodes, and if a parentheses is encountered, it will branch
//off into a new subtree and populate the node with data. The roots of the subtrees
//are operators, with digits (or a new operator possessing more children) for children.

void preOrderTraverse(BinaryTree *root);
//Precondition: root points to a populated binary expression tree
//Postcondition: Traverses the list recursively and outputs the prefix form of the
//infix expression

void postOrderTraverse(BinaryTree *root);
//Precondition: root points to a populated binary expression tree
//Postcondition: Traverses the list recursively and outputs the postfix
//form of the infix expression

void inOrderTraverse(BinaryTree *root);
//Precondition: root points to a populated binary expression tree
//Postcondition: Traverses the list recursively and outputs the infix
//form of the infix expression, but it will be in order (order of operations will not
//apply here since it's outputted in the order of evaluation).

double evaluate(BinaryTree *root);
//Precondition: root points to a populated binary expression tree
//Postcondition: The solution to the calculation of the data in the binary
//tree is returned recursively

bool isOperator(char ch);
//Precondition: ch has been given a value
//Postcondition: returns true if ch is either +, -, *, /, or ^

void changecolor(char* word, int color);
//precondition: word has been given a string value, and color has been given a number
//postcondition: conio function textcolor() has been called within function changecolor,
//changing the char* word into the color given by the int color.

void flushin();
//Postcondition: input buffer cleared to avoid unwanted input values


//===========================/=/=/=/=/=/=/=========================//
//                           IMPLEMENTATION                        //
//===========================/=/=/=/=/=/=/=========================//


int main(){
	char infix[50], yesOrNo; 
	int pos;                 //index variable
	BinaryTree *tree;        //our binary tree! Everyone say hi!
	
	do{
	
		clrscr();
		pos = 0;  //to be used as the index variable for infix. 0 is the start of a string
		changecolor("Binary Expression Tree\r\n\r\n", LIGHTGREEN);
		cout << "Give me a <fully> parenthesized infix equation: \n";
		changecolor("\r\nInfix Expression: ", WHITE); 
		cin.getline(infix, '\n');
		cout << endl << endl;
		
		tree = buildTree(infix, pos);      //inserts infix expression into a binary tree
		cout << "Prefix form of infix (preorder traversal): \n\t"; 
		preOrderTraverse(tree);   
		         
		cout << "\n\nPostfix form of infix (postorder traversal): \n\t"; 
		postOrderTraverse(tree);     
		     
		cout << "\n\nInfix form of infix (inorder traversal, read from left to right): \n\t"; 
		inOrderTraverse(tree);           //outputs infix form of infix, but from the tree! Amazing! 
		 
		changecolor("\r\n\r\nEvaluation of tree: ", LIGHTBLUE);
		cout << evaluate(tree);   //outputs the solution to the infix equation from the tree
		
		delete tree;   //returns the memory to the heap
		
		cout << "\n\n\nAgain? ";
		
		do{
			cin.get(yesOrNo);
		}while((tolower(yesOrNo) != 'y') && (tolower(yesOrNo) != 'n'));  //repeats if user doesn't enter y or n
		 
		flushin();   //clear input buffer
	}while(tolower(yesOrNo) == 'y');   //continues the program as long as they say 'y'
	
	return 0;
} //end main


/*struct BinaryTree function definitions*/


BinaryTree::BinaryTree(){   
	left = NULL;     
	right = NULL;
} //end BinaryTree default constructor


BinaryTree::~BinaryTree(){
		
	if (left != NULL)   //if the pointer is being used 
		delete left;    //return it to the heap
				
	if (left != NULL)   //if the pointer is being used 
		delete right;   //return it to the heap
				
} //end BinaryTree destructor


/*driver file function definitions*/


BinaryTree* buildTree(char *infix, int &pos){
	BinaryTree *temp_root = new BinaryTree;       //allocates memory
	
	pos++; //advance one character (either a ( or a digit)
	
	if (infix[pos] == '(')  //if you encounter a parenthese
		temp_root -> left = buildTree(infix, pos);    //recursive call: creates a subtree to the left
	
	else{
		temp_root -> left = new BinaryTree;             //allocate memory
		temp_root -> left -> data = infix[pos];         //assigns data unit
	}
	
	pos++;                        //so we are able to get the next character of infix
	temp_root -> data = infix[pos];    //assigns the operator to the root's data character
	pos++;                        //so we are able to get the next character of infix
	
	if (infix[pos] == '(')   //if a nested parenthese arises
		temp_root -> right = buildTree(infix, pos);   //recursive call: creates a subtree to the right!
	
	else{	
		temp_root -> right = new BinaryTree;           //allocate memory
		temp_root -> right -> data = infix[pos];       //assigns data unit
	}
	
	pos++;       //skip over the ending ) bracket
	return temp_root;
} //end buildTree()


void preOrderTraverse(BinaryTree *root){
	cout << (root -> data);			   //output the data of the root

	if (root -> left != NULL)
		preOrderTraverse(root -> left);  //traverse down to the left side of the tree
		
	if (root -> right != NULL)	  
		preOrderTraverse(root -> right); //traverse down to the right side of the tree	
		
} //end preOrderTraverse()


void postOrderTraverse(BinaryTree *root){	 

	if (root -> left != NULL)
		postOrderTraverse(root -> left);  //traverse down to the left side of the tree
	
	if (root -> left != NULL)		
		postOrderTraverse(root -> right); //traverse down to the right side of the tree
		
	cout << (root -> data);	      //output the data of the root
	
} //end postOrderTraverse()


void inOrderTraverse(BinaryTree *root){	   

	if (root -> left != NULL)
		inOrderTraverse(root -> left);  //traverse down to the left side of the tree
		
	cout << (root -> data);	      //output the data of the root	
	
	if (root -> left != NULL)		
		inOrderTraverse(root -> right); //traverse down to the right side of the tree
	
} //end inOrderTraverse()


double evaluate(BinaryTree *root){
	double temp, divisor;
	
	if (isdigit(root -> data))       //if the data character is a number
		temp = (root -> data) - 48;  //convert char to the respective int/double value (48 is the ascii offset)
		
		
	if (isOperator(root -> data)){   //if the data character is an operator
	
		switch (root -> data){   //apply the operator to the left and right values using recursive traversal
		
			case '+': 
				temp = evaluate(root -> left) + evaluate(root -> right); 
				break;
				
			case '-': 
				temp = evaluate(root -> left) - evaluate(root -> right); 
				break;
				
			case '*': 
				temp = evaluate(root -> left) * evaluate(root -> right); 
				break;
				
			case '/':
				divisor = evaluate(root -> right);   
				
				if (divisor == 0){   //in math, you cannot divide by 0
					cout << "\nDivide by 0 error.\n";
					exit(1);
				}
				
				else    //it's not 0, we can divide by it
					temp = evaluate(root -> left) / divisor;
					
				break;
				
			case '^': 
				temp = pow(evaluate(root -> left), evaluate(root -> right)); //raises the first arg to the second arg
				break;
				
			default: 
				cout << "\nError, operator unknown\n"; 
				break;	
			  
		}
	
	}	 
		
	return temp;
} //end evaluate()


bool isOperator(char ch){
	string acceptedOperators = "+-*/^";
	
	if(acceptedOperators.find(ch, 0) != string::npos)    //if ch can't be found in the string
		return true;
		
	else
		return false;
		
} //end isOperator()


void changecolor(char* word, int color){
	textcolor(color); 		 //changes text to the color given through the argument
	cprintf("%s", word);	   	    //outputs the string in the given color
	textcolor(LIGHTGRAY);	 //change color back to the default light gray
} //end changecolor()


void flushin(){

	while (cin.get() != '\n'){}   //clear the input buffer
	
} //end flushin()

My finished code.

It works for the most part. Unfortunately, it doesn't like equations like "(((3+4)*6)/7)" even though it should. Anyone able to spot why?

Nakeerb · 01-22-2003, 08:23 AM

It must be something small, this entire program works fairly well

01-21-2003, 12:28 AM	#2
Nakeerb Registered User Join Date: Oct 2002 Posts: 155	Alright there we go. Alignments fixed
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01-21-2003, 01:27 AM	#3
Nakeerb Registered User Join Date: Oct 2002 Posts: 155	Does this have anything to do with checking for left?
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01-21-2003, 01:44 AM	#4
Nick Blank Join Date: Aug 2001 Posts: 1,034	In each function you just look at the different cases that can arise and return a tree representing them. You only need to use the constructor node(char data, node* left, node* right). This way you return a tree even if it has incorrect data. In most of the functions all you need is node* root, right, left. So for example if your parsing 5 + 6 in expr you would create a tree for it by doing left = term(); right = expr(); root = new node('+', left, right); Be aware that if your just parsing 4; your function must basically do left = term(); root = left; This will go through the default case in expr and term. Last edited by Nick; 01-21-2003 at 01:47 AM.
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01-21-2003, 02:38 PM	#6
Nick Blank Join Date: Aug 2001 Posts: 1,034	His traversal is correct. As long as the tree is formed correctly it should run. But I think passing next_token into each function is ugly. Better to use a class or use a global variable. You should probably skip white space at the beginning of read_next_token and I'm not sure why you are using double c. Try to test each function in isolation. Simply test read_next_token then test factor, term, expr in that order.
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01-21-2003, 03:12 PM	#7
Bakamonoda Registered User Join Date: Jan 2003 Posts: 1	Nakeerb here, temporary name. Nick, then is all the new code I wrote a waste? I understand how you mean but I have a feeling I'd have to clean up SO much code to get it working. The ( and ) are not supposed to be part of the binary tree at all. only numbers and operators, and all evaluations are FULLY parenthesized. As in you'll never see (3 + 2 + (4 + 5)), it'd be ((3 + 2)+(4+5)) However, I have too much crappy code written in alerady. What would I do in place of something such as this: Code: switch(next_token.type){ case '#': //if the type of the token is a number val -> data = char(next_token.val); read_next_token(next_token); //get the next inputted letter break; case '(': //the only other thing it can be here is a ( read_next_token(next_token); //get the next inputted letter if (val -> left == NULL) //EQUAL TERMS YOU FOOL val -> left = expr(next_token); else val -> right = expr(next_token); read_next_token(next_token); //get the next inputted letter break; default: break; } If you can help me with this function I will do the rest. I am just totally lost. Do I need to declare two node pointers val and root? Or declare root when I... auugh! I couldn't sleep last night because of this stupid binary tree problem O.o;; when you say left, I assumed you meant a temp node's left is equal. Then you pass the temp node's left into the constructor... oyyy, no idea.
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01-21-2003, 06:19 PM	#9
Nakeerb Registered User Join Date: Oct 2002 Posts: 155	So I can do this for all functions? When would I need to append to right side or left?
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01-21-2003, 08:10 PM	#10
Nick Blank Join Date: Aug 2001 Posts: 1,034	This is just for factor. The functions like expr there are cases where you would set root = term() (this is in the default cases) and there are cases where you would do something like node* left = term(); node* right = expr(); root = node('+', left, right)
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01-22-2003, 08:23 AM	#14
Nakeerb Registered User Join Date: Oct 2002 Posts: 155	It must be something small, this entire program works fairly well
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