156. Binary Tree Upside Down

Binary Tree Upside Down

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.

For example: Given a binary tree {1,2,3,4,5},

1

/ \ 2 3 / \ 4 5

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

4 / \ 5 2 / \ 3 1

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public 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);
        TreeNode left = root.left;
        TreeNode right = root.right;

        left.left = right;
        left.right = root;
        root.left = null;
        root.right = null;
        return newRoot;
    }
}