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<int> out;
14 vector<vector<int>> res;
15 dfs(root,sum,0,out,res);
16 return res;
17 }
18 //还是递归先序遍历+vector<int>容器记录路径
19 void dfs(TreeNode* node,int sum,int curSum,vector<int> &out,vector<vector<int>> &res){
20 if(!node) return;
21 curSum+=node->val;
22 out.push_back(node->val);
23 if(node->left==NULL&&node->right==NULL&&curSum==sum) res.push_back(out);
24 dfs(node->left,sum,curSum,out,res);
25 dfs(node->right,sum,curSum,out,res);
26 out.pop_back();
27 }
28 };