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]
]

代码：
 1     vector<vector<int> > pathSum(TreeNode *root, int sum) {
 2         // IMPORTANT: Please reset any member data you declared, as
 3         // the same Solution instance will be reused for each test case.
 4         vector<vector<int> > result;
 5         result.clear();
 6         if(root == NULL)
 7             return result;
 8         if(root->val == sum && root->left == NULL && root->right == NULL){
 9             vector<int> tmp(1, sum);
10             result.push_back(tmp);
11             return result;
12         }
13         vector<vector<int> > lresult = pathSum(root->left, sum-root->val);
14         vector<vector<int> > rresult = pathSum(root->right, sum-root->val);
15         int n = lresult.size();
16         for(int i = 0; i < n; i++){
17             vector<int> tmp = lresult[i];
18             tmp.insert(tmp.begin(), root->val);
19             result.push_back(tmp);
20         }
21         n = rresult.size();
22         for(int i = 0; i < n; i++){
23             vector<int> tmp = rresult[i];
24             tmp.insert(tmp.begin(), root->val);
25             result.push_back(tmp);
26         }
27         return result;
28     }

 

posted on 2013-11-08 20:32  waruzhi  阅读(128)  评论(0)  编辑  收藏  举报

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