Leetcode 112. Path Sum

112. Path Sum

  • Total Accepted: 112241
  • Total Submissions: 352808
  • Difficulty: Easy

 

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.

 

思路：DFS。只要找到叶节点的时候，sum==0，就可以返回true；否则返回false。

 

代码：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool hasPathSum(TreeNode* root, int sum) {
13         if(!root) return false;
14         if(!root->left&&!root->right) return root->val==sum;//判断是否为叶节点
15         return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);
16     }
17 };

 

posted @ 2016-07-20 06:50  Deribs4  阅读(158)  评论(0)  编辑  收藏  举报