Maximum Depth of Binary Tree

Problem

Given the root of a binary tree, return its maximum depth.

Examples

The brute-force approach

There is no brute force here, recursion is already optimal. Alternatively, iterative BFS (level-order traversal) can count levels explicitly, but it uses O(n) extra space for the queue.

# BFS approach, counts levels
from collections import deque

if not root:
    return 0

depth = 0
queue = deque([root])

while queue:
    depth += 1
    for _ in range(len(queue)):   # process entire level
        node = queue.popleft()
        if node.left: queue.append(node.left)
        if node.right: queue.append(node.right)

return depth

BFS is O(n) time and O(n) space for the queue. Recursion is also O(n) time, but the call stack space is O(h) where h is the tree height, better for balanced trees.

Spotting the pattern

This is a Tree Traversal problem. The key question to ask yourself:

Can I compute the depth recursively as 1 + max(depth of left subtree, depth of right subtree)?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.