Binary Tree Upside Down — LeetCode 156 Python Solution
- Problem
- #156
- Pattern
- Tree Traversal
- Reading time
- 3 min
- Source
- leetcode.com
The problem
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: The original left child becomes the new root.
Example
- Input
- root = [1,2,3,4,5]
- Output
- [4,5,2,null,null,3,1]
Python solution
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def upsideDownBinaryTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if root is None or root.left is None:
return root
new_root = self.upsideDownBinaryTree(root.left)
root.left.right = root
root.left.left = root.right
root.left = None
root.right = None
return new_rootComplexity
| Measure | Complexity |
|---|---|
| Time | O(V+E) |
| Space | O(V) auxiliary |
Pattern: Tree Traversal
Choose the order — preorder, inorder, postorder, level — and the problem solves itself. LeetCode 156. Binary Tree Upside Down is filed here because LeetCode tags it Tree and Binary Tree, which is the vocabulary this hub collects.
The tree traversal guide has the Python template for the pattern and the 225 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 156. Binary Tree Upside Down?
- LeetCode 156. Binary Tree Upside Down is rated Medium on LeetCode.
- What is the time complexity of LeetCode 156. Binary Tree Upside Down?
- The Python solution on this page runs in O(V+E).
- What is the space complexity of LeetCode 156. Binary Tree Upside Down?
- The Python solution on this page uses O(V) auxiliary space.
- What topics does LeetCode 156. Binary Tree Upside Down cover?
- LeetCode 156. Binary Tree Upside Down is tagged Tree, Depth-First Search and Binary Tree on LeetCode.
- Is LeetCode 156. Binary Tree Upside Down a premium problem?
- Yes. LeetCode 156. Binary Tree Upside Down is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.