# 路径总和 II

<p>给你二叉树的根节点 <code>root</code> 和一个整数目标和 <code>targetSum</code> ，找出所有 <strong>从根节点到叶子节点</strong> 路径总和等于给定目标和的路径。</p>

<p><strong>叶子节点</strong> 是指没有子节点的节点。</p>

<div class="original__bRMd">
<div>
<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/abdc6911d9d9ee1b12dcdbb1945a1ab2.jpeg" style="width: 500px; height: 356px;" />
<pre>
<strong>输入：</strong>root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
<strong>输出：</strong>[[5,4,11,2],[5,8,4,5]]
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/9bc801134935f549d425954a120f1cc3.jpeg" style="width: 212px; height: 181px;" />
<pre>
<strong>输入：</strong>root = [1,2,3], targetSum = 5
<strong>输出：</strong>[]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>root = [1,2], targetSum = 0
<strong>输出：</strong>[]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>树中节点总数在范围 <code>[0, 5000]</code> 内</li>
	<li><code>-1000 <= Node.val <= 1000</code></li>
	<li><code>-1000 <= targetSum <= 1000</code></li>
</ul>
</div>
</div>


## template

```python

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        pathvalue = 0
        path = []
        result = []

        def preorder(node, pathvalue, sum, path, result):
            if node == None:
                return
            pathvalue += node.val
            path.append(node.val)
            if pathvalue == sum and node.left == None and node.right == None:
                result.append(list(path))  # 注意加list
            preorder(node.left, pathvalue, sum, path, result)
            preorder(node.right, pathvalue, sum, path, result)
            pathvalue -= node.val
            path.pop()

        preorder(root, pathvalue, sum, path, result)
        return result

```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```