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:

 bool hasPathSum(TreeNode* root, int sum) {
        if ( root == NULL )
        {
           return false;
        }
        
        if ( root->left == NULL && root->right == NULL)
            return root->val == sum;
        else
            return hasPathSum( root->left, sum - root->val ) || hasPathSum( root->right, sum - root->val );
    }