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Maximum Depth of Binary Tree

BFS — Level-by-Level Depth Counting

LeetCode 104 • Easy • Trees

Input: root = [3, 9, 20, null, null, 15, 7] → Output: 3
The maximum depth is the number of nodes along the longest path from root to the farthest leaf.

Time O(n) visit each node once

Space O(n) queue holds widest level

Visited: 0/5 Queue: 0 Depth: 0

Processing

In Queue

Processed

Edge highlight

Queue

empty

Depth

Speed:

Ready

Press Play to watch BFS count depth level-by-level, or Step to advance one operation at a time.
Each time we fully process a level, depth += 1.

✎ Whiteboard

Size: 3

⌨ Type It — Build Muscle Memory

Practice until you don't need to look. Use the guide comments below as scaffolding. Type the full implementation in the editor. The green highlights are the nuances to burn into memory.

═══ MAX DEPTH — BFS ITERATIVE ═══

PATTERN ▸ BFS + level counting O(n) · O(n)

① BASE CASE

if not root: return 0

② INIT

from collections import deque

queue = deque([root])

depth = 0

③ BFS LOOP

while queue:

level_size = len(queue)

depth += 1

④ PROCESS LEVEL

for i in range(level_size):

node = queue.popleft()

if node.left: queue.append(node.left)

if node.right: queue.append(node.right)

⑤ RETURN

return depth

▼ your implementation ▼

Verify your solution:

⌨ Type It — Recursive DFS

Practice until you don't need to look. The green highlights are the nuances to burn into memory.

═══ MAX DEPTH — RECURSIVE DFS ═══

PATTERN ▸ Post-order DFS O(n) · O(h)

① BASE CASE

if not root: return 0

② RECURSE BOTH SUBTREES

left = self.maxDepth(root.left)

right = self.maxDepth(root.right)

③ COMBINE

return 1 + max(left, right)

▼ your implementation ▼

Verify your solution: