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Path Sum II

时间:2014-09-13 20:05:35      阅读:219      评论:0      收藏:0      [点我收藏+]

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Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             /             4   8
           /   /           11  13  4
         /  \    /         7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

 1 /** 
 2  * Definition for binary tree 
 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     vector<vector<int> > pathSum(TreeNode *root, int sum) {  
13           
14         vector<vector<int> > gather;  
15         vector<int> rootLeaf;  
16         int tempSum = 0;  
17         pathSum(root, tempSum, sum, gather, rootLeaf);  
18         return gather;  
19     }  
20       
21     void pathSum(TreeNode *root, int tempSum, int sum, vector<vector<int> > &gather, vector<int> rootLeaf)  
22     {  
23         if(root != NULL)  
24         {  
25             if(root->left == NULL && root->right == NULL)  
26             {  
27                 tempSum += root->val;  
28                 rootLeaf.push_back(root->val);  
29                 if(tempSum == sum)  
30                     gather.push_back(rootLeaf);  
31                   
32                 return;  
33             }  
34           
35             else  
36             {  
37                 tempSum += root->val;  
38                 rootLeaf.push_back(root->val);  
39                 pathSum(root->left, tempSum, sum, gather, rootLeaf);  
40                 pathSum(root->right, tempSum, sum, gather, rootLeaf);  
41             }  
42         }  
43     }  
44 };  

 

Path Sum II

标签:style   blog   color   io   ar   for   div   sp   log   

原文地址:http://www.cnblogs.com/YQCblog/p/3970189.html

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