leecode/problems/113.path-sum-ii.md

## 题目地址
https://leetcode.com/problems/path-sum-ii/description/

## 题目描述
```
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

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]
]
```

## 思路

这道题目是求集合，并不是`求值`，而是枚举所有可能，因此动态规划不是特别切合，因此我们需要考虑别的方法。

这种题目其实有一个通用的解法，就是回溯法。
网上也有大神给出了这种回溯法解题的
[通用写法](https://leetcode.com/problems/combination-sum/discuss/16502/A-general-approach-to-backtracking-questions-in-Java-(Subsets-Permutations-Combination-Sum-Palindrome-Partitioning))，这里的所有的解法使用通用方法解答。
除了这道题目还有很多其他题目可以用这种通用解法，具体的题目见后方相关题目部分。

我们先来看下通用解法的解题思路，我画了一张图：

![backtrack](../assets/problems/backtrack.png)

通用写法的具体代码见下方代码区。

## 关键点解析

- 回溯法
- backtrack 解题公式


## 代码

* 语言支持：JS，C++，Python3

JavaScript Code：

```js
/*
 * @lc app=leetcode id=113 lang=javascript
 *
 * [113] Path Sum II
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
function backtrack(root, sum, res, tempList) {
  if (root === null) return;
  if (root.left === null && root.right === null && sum === root.val)
    return res.push([...tempList, root.val]);

  tempList.push(root.val);
  backtrack(root.left, sum - root.val, res, tempList);

  backtrack(root.right, sum - root.val, res, tempList);
  tempList.pop();
}
/**
 * @param {TreeNode} root
 * @param {number} sum
 * @return {number[][]}
 */
var pathSum = function(root, sum) {
  if (root === null) return [];
  const res = [];
  backtrack(root, sum, res, []);
  return res;
};
```
C++ Code:
```C++
/**
 * Definition for a binary tree node.
 * 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) {
        auto ret = vector<vector<int>>();
        auto temp = vector<int>();
        backtrack(root, sum, ret, temp);
        return ret;
    }
private:
    void backtrack(const TreeNode* root, int sum, vector<vector<int>>& ret, vector<int>& tempList) {
        if (root == nullptr) return;
        tempList.push_back(root->val);
        if (root->val == sum && root->left == nullptr && root->right == nullptr) {
            ret.push_back(tempList);
        } else {
            backtrack(root->left, sum - root->val, ret, tempList);
            backtrack(root->right, sum - root->val, ret, tempList);
        }
        tempList.pop_back();        
    }
};
```
```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        if not root:
            return []
        
        result = []
        
        def trace_node(pre_list, left_sum, node):
            new_list = pre_list.copy()
            new_list.append(node.val)
            if not node.left and not node.right:
                # 这个判断可以和上面的合并，但分开写会快几毫秒，可以省去一些不必要的判断
                if left_sum == node.val:
                    result.append(new_list)
            else:
                if node.left:
                    trace_node(new_list, left_sum-node.val, node.left)
                if node.right:
                    trace_node(new_list, left_sum-node.val, node.right)
        
        trace_node([], sum, root)
        return result
```
## 相关题目

- [39.combination-sum](./39.combination-sum.md)
- [40.combination-sum-ii](./40.combination-sum-ii.md)
- [46.permutations](./46.permutations.md)
- [47.permutations-ii](./47.permutations-ii.md)
- [78.subsets](./78.subsets.md)
- [90.subsets-ii](./90.subsets-ii.md)
- [131.palindrome-partitioning](./131.palindrome-partitioning.md)