Path Sum

Description:

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5

             / \

            4   8

           /   / \

          11  13  4

         /  \      \

        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Code:

 1  bool hasPathSum(TreeNode* root, int sum) {

 2         if ( root == NULL )

 3         {

 4            return false;

 5         }

 6         

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

 8             return root->val == sum;

 9         else

10             return hasPathSum( root->left, sum - root->val ) || hasPathSum( root->right, sum - root->val );

11     }