# LeetCode 814. Binary Tree Pruning

## Description

https://leetcode.com/problems/matrix-block-sum/

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:

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

Example 2:

```Input: root = [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]
```

Example 3:

```Input: root = [1,1,0,1,1,0,1,0]
Output: [1,1,0,1,1,null,1]
```

Constraints:

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

## Explanation

Check the subtree recursively, if the node value is 1, or the node contains subrees with value 1, keep the subtree, otherwise, make the node null.

## Python Solution

``````# 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 self.helper(root):
return None

return root

def helper(self, root):
if not root:
return False

left = self.helper(root.left)
right = self.helper(root.right)

if not left:
root.left = None

if not right:
root.right = None

return root.val == 1 or left or right``````
• Time Complexity: O(N).
• Space Complexity: O(N).