Path Sum

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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.

 1     bool hasPathSum(TreeNode* root, int sum) {
 2         if(root == NULL)
 3             return false;
 4         
 5         sum -= root->val;
 6         if(root->left == NULL && root->right == NULL){
 7             return sum == 0;
 8         }
 9         
10         return (hasPathSum(root->left, sum) || hasPathSum(root->right, sum));
11     }

Path Sum

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原文地址：http://www.cnblogs.com/horizonice/p/4770550.html