Tuesday 14 June 2011

Counting number of nodes in a binary tree



   How would you count the number of nodes in a tree? Traverse all the nodes in a tree. You can do any traversal and just start counting. But there is a concept called tail recursion which is handy and very easily readable.

If you have a tree, if you know the number of nodes in the left and number of nodes in right +1, then, what you get is total number of nodes.


   You have to do tail recursion with count(root->left) + count(root->right)+1.

After you get the count, how you will find the height of the tree ?

  log2 count_of_nodes. But in C++, you won’t get a function to calculate log base 2. :)

There is a way,

  log10 n/log 10 2 = log 2 n

So, you just need to use the above formula to get log base 2 which yields the height of the tree.


Tree* tobj;
int count(Tree* root) {
    if(root == NULL)
        return 0;
        if(root->left == NULL && root->right == NULL)
            return 1;
            return count(root->left) + count(root->right) + 1;    
  • you are covering 2 base conditions here, root == NULL then 0. root->left & root->right is NULL, then only root node is there, which is 1.
  • This expands to 1 for each node in the left & right.


An2quamaraN said...

One of my exam questions was to write this algorithm. I used your version. We'll see how good it is ;)

Unknown said...

The Condition:

if(root->left == NULL && root->right == NULL)

return 1;

is NOT needed, if both Children are NULL, we will get 0 for them.

(root==NULL) condition is enough to serve as the base condition.

vprajan said...


Yes Sandeep. you are right!! the extra condition is just to reduce the extra function calls.

Unknown said...

Thanks a lot great tips,very useful by quizvook.com

Unknown said...

I am trying to figure out how to call this function in main. :(

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