# [814. 二叉树剪枝](https://leetcode-cn.com/problems/binary-tree-pruning)

[English Version](/solution/0800-0899/0814.Binary%20Tree%20Pruning/README_EN.md)

## 题目描述

<!-- 这里写题目描述 -->

<p>给定二叉树根结点&nbsp;<code>root</code>&nbsp;，此外树的每个结点的值要么是 0，要么是 1。</p>

<p>返回移除了所有不包含 1 的子树的原二叉树。</p>

<p>( 节点 X 的子树为 X 本身，以及所有 X 的后代。)</p>

<pre>
<strong>示例1:</strong>
<strong>输入:</strong> [1,null,0,0,1]
<strong>输出: </strong>[1,null,0,null,1]

<strong>解释:</strong>
只有红色节点满足条件&ldquo;所有不包含 1 的子树&rdquo;。
右图为返回的答案。

<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0814.Binary%20Tree%20Pruning/images/1028_2.png" style="width:450px" />
</pre>

<pre>
<strong>示例2:</strong>
<strong>输入:</strong> [1,0,1,0,0,0,1]
<strong>输出: </strong>[1,null,1,null,1]


<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0814.Binary%20Tree%20Pruning/images/1028_1.png" style="width:450px" />
</pre>

<pre>
<strong>示例3:</strong>
<strong>输入:</strong> [1,1,0,1,1,0,1,0]
<strong>输出: </strong>[1,1,0,1,1,null,1]


<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0814.Binary%20Tree%20Pruning/images/1028.png" style="width:450px" />
</pre>

<p><strong>说明: </strong></p>

<ul>
	<li>给定的二叉树最多有&nbsp;<code>100</code>&nbsp;个节点。</li>
	<li>每个节点的值只会为&nbsp;<code>0</code> 或&nbsp;<code>1</code>&nbsp;。</li>
</ul>

## 解法

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### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pruneTree(self, root: TreeNode) -> TreeNode:
        if not root:
            return None
        root.left = self.pruneTree(root.left)
        root.right = self.pruneTree(root.right)
        if root.val == 0 and not root.left and not root.right:
            return None
        return root
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode pruneTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        root.left = pruneTree(root.left);
        root.right = pruneTree(root.right);
        if (root.val == 0 && root.left == null && root.right == null) {
            return null;
        }
        return root;
    }
}
```

### **Go**

```go
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pruneTree(root *TreeNode) *TreeNode {
	if root == nil {
		return nil
	}
	root.Left = pruneTree(root.Left)
	root.Right = pruneTree(root.Right)
	if root.Val == 0 && root.Left == nil && root.Right == nil {
		return nil
	}
	return root
}
```

### **C++**

```cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* pruneTree(TreeNode* root) {
        if (!root) return nullptr;
        root->left = pruneTree(root->left);
        root->right = pruneTree(root->right);
        if (!root->val && !root->left && !root->right) return nullptr;
        return root;
    }
};
```

### **...**

```

```

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