【Leetcode】【Medium】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]
]

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：

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     vector<vector<int> > pathSum(TreeNode *root, int sum) {
         vector<vector<int> > ret;
         vector<int> cur_nums;
         pathSumTree(ret, cur_nums, root, sum);
         return ret;
     }

     void pathSumTree(vector<vector<int> > &ret, vector<int> cur_nums, TreeNode *root, int cur_sum) {
         if (!root)
             return;

         cur_nums.push_back(root->val);
         if (!root->left && !root->right && cur_sum - root->val == )
             ret.push_back(cur_nums);

         pathSumTree(ret, cur_nums, root->left, cur_sum - root->val);
         pathSumTree(ret, cur_nums, root->right, cur_sum - root->val);
         return;
     }
 };

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：

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     vector<vector<int> > pathSum(TreeNode *root, int sum) {
         vector<vector<int> > ret;
         vector<int> cur_nums();
         pathSumTree(ret, cur_nums, root, sum);
         return ret;
     }

     void pathSumTree(vector<vector<int> > &ret, vector<int> &cur_nums, TreeNode *root, int cur_sum) {
         if (!root)
             return;

         cur_nums.push_back(root->val);
         if (!root->left && !root->right && cur_sum - root->val == ) {
             ret.push_back(cur_nums);
             cur_nums.pop_back();
             return;
         }

         pathSumTree(ret, cur_nums, root->left, cur_sum - root->val);
         pathSumTree(ret, cur_nums, root->right, cur_sum - root->val);
         cur_nums.pop_back();
         return;
     }
 };