112. 路径总和 

 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 {
12 public:
13     bool hasPathSum(TreeNode* root, int sum) 
14     {
15         //如果root为空，直接返回false
16         if(root == NULL) return false;
17         //如果左、右孩子都为空且sum为当前节点的值，返回true
18         if(root->left == NULL && root->right == NULL && sum == root->val) return true;
19         return hasPathSum(root->left,sum - root->val) 
20             || hasPathSum(root->right,sum - root->val);
21     }
22 };

 

posted @ 2020-04-01 15:18  Jinxiaobo0509  阅读(103)  评论(0)    收藏  举报