https://github.com/emahtab/binary-tree-pruning

Binary Tree Pruning
https://github.com/emahtab/binary-tree-pruning

binary-tree leetcode problem-solving pruning

Last synced: 4 months ago
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Binary Tree Pruning

• Host: GitHub
• URL: https://github.com/emahtab/binary-tree-pruning
• Owner: eMahtab
• Created: 2020-02-02T12:33:10.000Z (over 5 years ago)
• Default Branch: master
• Last Pushed: 2020-03-11T17:55:24.000Z (over 5 years ago)
• Last Synced: 2025-02-02T03:26:16.477Z (9 months ago)
• Topics: binary-tree, leetcode, problem-solving, pruning
• Homepage:
• Size: 23.4 KB
• Stars: 1
• Watchers: 1
• Forks: 0
• Open Issues: 0
• Metadata Files:
  • Readme: README.md

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README

          
# Binary Tree Pruning

## https://leetcode.com/problems/binary-tree-pruning

We are given the head node root of a binary tree, where additionally every node's value is either a 0 or a 1.

Return the same tree where every subtree (of the given tree) not containing a 1 has been removed.

(Recall that the subtree of a node X is X, plus every node that is a descendant of X.)

**Note:**

1. The binary tree will have at most 100 nodes.

2. The value of each node will only be 0 or 1.

![Binary Tree Pruning](binary-tree-pruning.PNG?raw=true "Binary Tree Pruning")

# Implementation :

```java

/**

 * Definition for a binary tree node.

 * public class TreeNode {

 *     int val;

 *     TreeNode left;

 *     TreeNode right;

 *     TreeNode(int x) { val = x; }

 * }

 */

class Solution {

    public TreeNode pruneTree(TreeNode root) {

        if(root == null)

            return null;

        return prune(root);

    }

    

    public TreeNode prune(TreeNode node){

        if(node == null)

            return node;

        node.left = prune(node.left);

        node.right = prune(node.right);

        return (node.left == null && node.right == null && node.val == 0) ?

                null : node;

    }

}

```

# Key Points :

Note that in the recursive prune method we are first calling the prune on left subtree and right subtree, if we move the check `(node.val == 0 && node.left == null && node.right == null)` before calling prune on left subtree and right subtree, it will lead to wrong result.

# References :

https://www.youtube.com/watch?v=77LJc56bwnE