题目：
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] ]
解题思路：
简单的DFS。
代码：
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 11 class Solution { 12 public: 13 int vec_sum(vector<int> &cur_ans) { 14 int sum = 0; 15 for (int i = 0; i < cur_ans.size(); i++) { 16 sum += cur_ans[i]; 17 } 18 return sum; 19 } 20 void search_dfs(TreeNode *root, vector<int> &cur_ans, int sum, vector<vector<int>> &ans) { 21 if (root->left == NULL && root->right == NULL) { 22 cur_ans.push_back(root->val); 23 if (vec_sum(cur_ans) == sum) ans.push_back(cur_ans); 24 cur_ans.pop_back(); 25 } 26 else { 27 cur_ans.push_back(root->val); 28 if (root->left != NULL) search_dfs(root->left, cur_ans, sum, ans); 29 if (root->right != NULL) search_dfs(root->right, cur_ans, sum, ans); 30 cur_ans.pop_back(); 31 } 32 return; 33 } 34 vector<vector<int> > pathSum(TreeNode *root, int sum) { 35 vector<vector<int>> ans; 36 if (root == NULL) return ans; 37 38 vector<int> cur_ans; 39 40 search_dfs(root, cur_ans, sum, ans); 41 return ans; 42 } 43 };