156. Binary Tree Upside Down

Description

Given the root of a binary tree, turn the tree upside down and return the new root.

You can turn a binary tree upside down with the following steps:

1. The original left child becomes the new root.

2. The original root becomes the new right child.

3. The original right child becomes the new left child.

The mentioned steps are done level by level, it is guaranteed that every node in the given tree has either 0 or 2 children.

Constraints

• The number of nodes in the tree will be in the range [0, 10].

• 1 <= Node.val <= 10

• Every node has either 0 or 2 children.

Approach

Links

• GeeksforGeeks
• Leetcode
• ProgramCreek

• YouTube

Examples

Example 1

Input: root = [1, 2, 3, 4, 5]

Output: [4, 5, 2, null, null, 3, 1]

Example 2

Input: root = []

Output: []

Example 3

Input: root = [1]

Output: [1]

Solutions

TreeNode

// 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;
    }
}

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        if(root == null) return null;
        
        if(root.left == null && root.right == null) return root;
        
        TreeNode newRoot = upsideDownBinaryTree(root.left);
        
        root.left.left = root.right;
        root.left.right = root;
        root.left = null;
        root.right = null;
        
        return newRoot;
    }
}

Solution 2

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        TreeNode curr = root;
        TreeNode next = null;
        TreeNode prev = null;
        TreeNode temp = null;
        
        while(curr != null) {
            next = curr.left;
            curr.left = temp;
            temp = curr.right;
            curr.right = prev;
            prev = curr;
            curr = next;
        }
        
        return prev;
    }
}

Follow up