# 113. Path Sum II

Medium

***

Given the `root` of a binary tree and an integer `targetSum`, return *all **root-to-leaf** paths where the sum of the node values in the path equals* `targetSum`*. Each path should be returned as a list of the node **values**, not node references*.

A **root-to-leaf** path is a path starting from the root and ending at any leaf node. A **leaf** is a node with no children.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/01/18/pathsumii1.jpg)

<pre><code>Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
<strong>Output:
</strong> [[5,4,11,2],[5,8,4,5]]
<strong>Explanation:
</strong> There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22
</code></pre>

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg)

<pre><code>Input: root = [1,2,3], targetSum = 5
<strong>Output:
</strong> []
</code></pre>

**Example 3:**

<pre><code>Input: root = [1,2], targetSum = 0
<strong>Output:
</strong> []
</code></pre>

&#x20;

**Constraints:**

* The number of nodes in the tree is in the range `[0, 5000]`.
* `-1000 <= Node.val <= 1000`
* `-1000 <= targetSum <= 1000`

#### Solution 1

```python
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        self.target = targetSum
        self.result = []
        self.dfs(root, 0 , [])
        return self.result
    
    def dfs(self, node, total = 0, path = []):
        if not node:
            return
        total += node.val
        path.append(node.val)
        if total == self.target and not node.left and not node.right:
            self.result.append(path)
            return
        self.dfs(node.left, total, path[::])
        self.dfs(node.right, total, path[::])
```

#### Solution 2

```python
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        result=[]
        def recursion(root, current , target, result):
            # print(root.val)
            if not root:
                return
            if not root.left and not root.right and target - (root.val) == 0:
                result.append(current + [root.val])
                return
            recursion(root.left, current + [root.val], target-root.val, result)
            recursion(root.right, current + [root.val], target-root.val, result)
        recursion(root, [] , targetSum, result)
        return resultpy
```