# 814. Binary Tree Pruning
### Description
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`.
### Constraints
* The number of nodes in the tree is in the range `[1, 200]`.
* `Node.val` is either `0` or `1`.
### Approach
### Links
* Binarysearch
* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/binary-tree-pruning/)
* ProgramCreek
* YouTube
### **Examples**
{% tabs %}
{% tab title="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.
{% endtab %}
{% tab title="Example 2" %}
**Input:** root = \[1, 0, 1, 0, 0, 0, 1]
**Output:** \[1, null, 1, null, 1]
**Explanation:**
{% endtab %}
{% tab title="Example 3" %}
**Input:** root = \[1, 1, 0, 1, 1, 0, 1, 0]
**Output:** \[1, 1, 0, 1, 1, null, 1]
**Explanation:**
{% endtab %}
{% endtabs %}
### **Solutions**
{% tabs %}
{% tab title="TreeNode" %}
```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;
}
}
```
{% endtab %}
{% tab title="Solution 1" %}
```java
/**
* Time complexity :
* Space complexity :
*/
class Solution {
public TreeNode pruneTree(TreeNode root) {
if(root == null) {
return null;
}
root.left = pruneTree(root.left);
root.right = pruneTree(root.right);
if(root.left != null || root.right != null || root.val == 1) {
return root;
}
return null;
}
}
```
{% endtab %}
{% endtabs %}
### **Follow up**
*
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