Algorithm Notes
Summary: Invert Binary Tree — notes not yet curated.
Time: Estimate via loops/recurrences; common classes: O(1), O(log n), O(n), O(n log n), O(n^2)
Space: Count auxiliary structures and recursion depth.
Tip: See the Big-O Guide for how to derive bounds and compare trade-offs.
Big-O Guide
Source
"""
Invert Binary Tree
TODO: Add problem description
"""
from src.interview_workbook.leetcode._registry import register_problem
from src.interview_workbook.leetcode._types import Category, Difficulty
class Solution:
def solve(self, root):
"""
Invert a binary tree (mirror it).
"""
if not root:
return None
root.left, root.right = self.solve(root.right), self.solve(root.left)
return root
def demo():
"""Run a simple demonstration of invert binary tree with BFS before/after."""
import random
random.seed(0)
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
# Build tree:
# 4
# / \
# 2 7
# / \ / \
# 1 3 6 9
root = TreeNode(4)
root.left = TreeNode(2)
root.right = TreeNode(7)
root.left.left = TreeNode(1)
root.left.right = TreeNode(3)
root.right.left = TreeNode(6)
root.right.right = TreeNode(9)
def bfs(node):
if not node:
return []
res = []
queue = [node]
while queue:
curr = queue.pop(0)
res.append(curr.val)
if curr.left:
queue.append(curr.left)
if curr.right:
queue.append(curr.right)
return res
before = bfs(root)
Solution().solve(root)
after = bfs(root)
return f"Before: {before}, After: {after}"
register_problem(
id=226,
slug="invert_binary_tree",
title="Invert Binary Tree",
category=Category.TREES,
difficulty=Difficulty.EASY,
tags=["tree", "binary_tree", "dfs", "bfs"],
url="https://leetcode.com/problems/invert-binary-tree/",
notes="",
)