Path Sum II

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]
]

 

Analyse: recursively do the left and right part. When a fraction meets the requirement, push it into the result.

Runtime: 12ms.

 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     vector<vector<int> > pathSum(TreeNode* root, int sum) {
13         vector<vector<int> > result;
14         if(!root) return result;
15         
16         vector<int> temp;
17         path(root, sum, result, temp);
18         return result;
19     }
20     void path(TreeNode* root, int gap, vector<vector<int> >& result, vector<int>& temp){
21         if(!root) return;
22         
23         temp.push_back(root->val);
24         if(!root->left && !root->right){
25             if(root->val == gap) result.push_back(temp);
26         }
27         path(root->left, gap - root->val, result, temp);
28         path(root->right, gap - root->val, result, temp);
29         
30         temp.pop_back();
31     }
32 };

 

posted @ 2015-08-01 23:06  amazingzoe  阅读(151)  评论(0)    收藏  举报