> For the complete documentation index, see [llms.txt](https://leetcode-solution-leetcode-pp.gitbook.io/leetcode-solution/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://leetcode-solution-leetcode-pp.gitbook.io/leetcode-solution/medium/113.path-sum-ii.md).

# 0113. 路径总和 II

### 题目地址(113. 路径总和 II)

<https://leetcode-cn.com/problems/path-sum-ii/>

### 题目描述

```
给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

示例:
给定如下二叉树，以及目标和 sum = 22，

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1
返回:

[
   [5,4,11,2],
   [5,8,4,5]
]

```

### 前置知识

* 回溯法

### 公司

* 阿里
* 腾讯
* 百度
* 字节

### 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：

![](https://p.ipic.vip/m71dgr.jpg)

> 图是 [78.subsets](/leetcode-solution/medium/78.subsets.md)，都差不多，仅做参考。

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:

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