Path Sum

来源:互联网 发布:点播软件 编辑:程序博客网 时间:2024/05/31 04:03

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.

class Solution {public:bool judge(TreeNode* root, int last, int sum){if (0 == root) return false;if ( 0 == root->left && 0 == root->right)return (root->val + last == sum) ? true:false;bool res = judge(root->left, last + root->val, sum);return res? res:judge(root->right, last + root->val, sum);}    bool hasPathSum(TreeNode *root, int sum){if (root == 0) return false;        return judge(root, 0, sum);    }};



0 0
原创粉丝点击