Path Sum

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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     }

Path Sum

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原文地址：http://www.cnblogs.com/happygirl-zjj/p/4590636.html