LeetCode 156. Binary Tree Upside Down

156. Binary Tree Upside Down

Difficulty: Medium

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

Example:

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

    1
   / \
  2   3
 / \
4   5

Output: return the root of the binary tree [4,5,2,#,#,3,1]

   4
  / \
 5   2
    / \
   3   1

Clarification:

Confused what [4,5,2,#,#,3,1] means? Read more below on how binary tree is serialized on OJ.

The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:

  1
 / \
2   3
   /
  4
   \
    5

The above binary tree is serialized as [1,2,3,#,#,4,#,#,5].

Solution

Language: 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 upsideDownBinaryTree(TreeNode root) {
        if (root == null) {
            return root;
        }
        if (root.left != null) {
            return helper(root.left, root);
        }
        return root;
    }
    
    private TreeNode helper(TreeNode left, TreeNode root) {
        TreeNode leftRoot;
        if (left.left != null) {
            leftRoot = helper(left.left, left);
        } else {
            leftRoot = left;
        }
        left.right = root;
        left.left = root.right;
        root.left = null;
        root.right = null;
        return leftRoot;
    }
}