# 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

```java
class Solution {
	public List<List<Integer>> pathSum(TreeNode root, int sum) {
		if (root == null)
			return new ArrayList<>();
		List<List<Integer>> ans = new ArrayList<>();
		if (root.val == sum && root.left == null && root.right == null) {
			List<Integer> arr = new ArrayList<>();
			arr.add(root.val);
			ans.add(arr);
			return ans;
		}
		List<List<Integer>> left = pathSum(root.left, sum - root.val);
		List<List<Integer>> right = pathSum(root.right, sum - root.val);
		for (List<Integer> list : left) {

			list.add(0, root.val);
			ans.add(list);
		}
		for (List<Integer> list : right) {
			list.add(0, root.val);
			ans.add(list);
		}
		return ans;
	}
}


```

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```