156

156. Binary Tree Upside Down (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].

Companies:
LinkedIn

Related Topics:
Tree

Similar Questions:

• Reverse Linked List (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/binary-tree-upside-down/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    TreeNode* upsideDownBinaryTree(TreeNode* root) {
        if (!root || !root->left) return root;
        TreeNode *left = root->left, *right = root->right, *newRoot = NULL;
        while (left) {
            TreeNode *nextLeft = left->left, *nextRight = left->right;
            if (!newRoot) root->left = root->right = NULL;
            newRoot = left;
            newRoot->left = right;
            newRoot->right = root;
            root = left;
            left = nextLeft;
            right = nextRight;
        }
        return newRoot;
    }
};