binary tree depth

hi,
i just want to count the levels of random binary tree.
i have used a queue to do a level-order traversal.
So the depth is equal to the height of the tree node
stored int the tail of the queue,
here's the algorithms i used to get the level of the tree

Code:

int levelCtr, nodeCtr = 0;
	QueueNodePtr headPtr = NULL, tailPtr = NULL, indexPtr = NULL, lastPtr = NULL;
	TreeNodePtr p;/*more code*/
lastPtr = tailPtr;
	indexPtr = headPtr;
	
	p = &( lastPtr -> qData );
	
	while ( lastPtr != headPtr ) {
		while ( ( indexPtr -> qData.leftPtr != &( lastPtr -> qData ) ) || ( indexPtr -> qData.rightPtr != &( lastPtr -> qData ) ) ) {
			indexPtr = indexPtr -> nextPtr;
		}
		
		lastPtr = indexPtr;
		indexPtr = headPtr;
		
		levelCtr++;
	}

but the debugger shows that p doesn't get the address of qData,
but the address of lastPtr

any hints are welcome,
trk

Code:

int depthCount( Node *n )
{
    int x=1,y=1;

    if ( !n ) return 0;

    x += depthCount( n->left );
    y += depthCount( n->right );
    return x > y ? x : y;
}

I think that'll do the trick. I haven't tested it, I just threw it together. However, it should work for what you want, assuming I'm understanding what you want done.

Quzah.

but the debugger shows that p doesn't get the address of qData,
but the address of lastPtr

This is normal. The address of the first element of a structure is generally the address of the structure.

Code:

#include <stdio.h>

int main (void)
{
 struct
 { int x, y} myStruct;
 printf ("myStruct: %p\n", &myStruct);
 printf ("myStruct.x: %p\n", &myStruct.x);
 printf ("myStruct.y: %p\n", &myStruct.y);

 return 0;
}

And the output

Code:

myStruct: 8f120
myStruct.x: 8f120
myStruct.y: 8f124

Quzah's algorithm is the correct(TM) way to go about this kind of problem though; figure out the depth of the left subtree, the right subtree, pick the biggest, and add 1.

The members of a structure have
the address of the structure...i got that now.
I was just trying to find another way except the usual
recursive function.

thanks and sorry for replying late,
it's been a busy week