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# 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.

<div align="left"><img src="/files/-MfI0hx21bT-juS9iOIG" alt=""></div>
{% endtab %}

{% tab title="Example 2" %}
**Input:** root = \[1, 0, 1, 0, 0, 0, 1]

**Output:** \[1, null, 1, null, 1]

**Explanation:**

<div align="left"><img src="/files/-MfI0n2RJMr6u8nNq62h" alt=""></div>
{% endtab %}

{% tab title="Example 3" %}
**Input:** root = \[1, 1, 0, 1, 1, 0, 1, 0]

**Output:** \[1, 1, 0, 1, 1, null, 1]

**Explanation:**

<div align="left"><img src="/files/-MfI0yzZBHvti-XG3ZwK" alt=""></div>
{% 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**

*