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Problem Statement
You are given a binary tree, return its level order traversal.

My thoughts
- Use BFS with a queue
- Track levels by queue length

Pseudocode
queue = [root]
while queue:
  for i in range(len(queue)):
    node = queue.pop(0)
    push node.val
    push children

Solution (Python)
# BFS level order traversal
# Time: O(n), Space: O(n)
from collections import deque

def level_order(root):
    if not root: return []
    res, q = [], deque([root])
    while q:
        level = []
        for _ in range(len(q)):
            n = q.popleft()
            level.append(n.val)  # capture value
            if n.left: q.append(n.left)
            if n.right: q.append(n.right)
        res.append(level)
    return res