# 814. Binary Tree Pruning

#### Medium

***

Given the `root` of a binary tree, return *the same tree where every subtree (of the given tree) not containing a* `1` *has been removed*.

A subtree of a node `node` is `node` plus every node that is a descendant of `node`.

 

**Example 1:**

![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/06/1028_2.png)

<pre><code>Input: root = [1,null,0,0,1]
<strong>Output:
</strong> [1,null,0,null,1]
<strong>Explanation:
</strong> 
Only the red nodes satisfy the property "every subtree not containing a 1".
The diagram on the right represents the answer.
</code></pre>

**Example 2:**

![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/06/1028_1.png)

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

**Example 3:**

![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/05/1028.png)

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

&#x20;

**Constraints:**

* The number of nodes in the tree is in the range `[1, 200]`.
* `Node.val` is either `0` or `1`.

```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: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        left = self.pruneTree(root.left)
        right = self.pruneTree(root.right)
        if not left:
            root.left = None
        if not right:
            root.right = None
        return root if left or right or root.val == 1 else None
```