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：
实用递归的方法，题目很简单，每深一层递归，带入新计算出的需要求的sum值，递归函数需要四个入参：最终返回数组，当前已经经历的路径，当前需要的sum，已经结点指针
由于需要递归多个函数，刚开始为了防止“当前已经经历的路径”出现重复记载，因而没有形参没有使用指针，而完整的传入了整个数组，时间72ms
代码如下：
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> > ret; 14 vector<int> cur_nums; 15 pathSumTree(ret, cur_nums, root, sum); 16 return ret; 17 } 18 19 void pathSumTree(vector<vector<int> > &ret, vector<int> cur_nums, TreeNode *root, int cur_sum) { 20 if (!root) 21 return; 22 23 cur_nums.push_back(root->val); 24 if (!root->left && !root->right && cur_sum - root->val == 0) 25 ret.push_back(cur_nums); 26 27 pathSumTree(ret, cur_nums, root->left, cur_sum - root->val); 28 pathSumTree(ret, cur_nums, root->right, cur_sum - root->val); 29 return; 30 } 31 };
但是将形参设置为指针后，代码运行时间大大减少了，因为传递指针比传递整个数组高效多了；
为了防止出现混乱，可以在每次路径考察结束后，将当前的结点从数组中pop出来，避免重复；
代码如下，这次只需要不到20ms：
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> > ret; 14 vector<int> cur_nums(0); 15 pathSumTree(ret, cur_nums, root, sum); 16 return ret; 17 } 18 19 void pathSumTree(vector<vector<int> > &ret, vector<int> &cur_nums, TreeNode *root, int cur_sum) { 20 if (!root) 21 return; 22 23 cur_nums.push_back(root->val); 24 if (!root->left && !root->right && cur_sum - root->val == 0) { 25 ret.push_back(cur_nums); 26 cur_nums.pop_back(); 27 return; 28 } 29 30 pathSumTree(ret, cur_nums, root->left, cur_sum - root->val); 31 pathSumTree(ret, cur_nums, root->right, cur_sum - root->val); 32 cur_nums.pop_back(); 33 return; 34 } 35 };