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]
]

Solution: DFS.

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

 

转载于:https://www.cnblogs.com/zhengjiankang/p/3649758.html