首页 > 其他 > 详细

Path Sum

时间:2014-05-30 15:19:11      阅读:300      评论:0      收藏:0      [点我收藏+]

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.

bubuko.com,布布扣
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if(root==NULL) return false;
        if(root->val==sum && root->left==NULL && root->right==NULL) return true;
        
        return hasPathSum(root->left,sum-root->val) || hasPathSum(root->right,sum-root->val);
    }
};
bubuko.com,布布扣

 

Path Sum,布布扣,bubuko.com

Path Sum

原文:http://www.cnblogs.com/erictanghu/p/3759630.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!