Binary Tree Level Order Traversal

Problem

Given the root of a binary tree, return the level order traversal of its nodes' values (i.e., from left to right, level by level).

Examples

The brute-force approach

DFS with a depth parameter — recurse, passing the current depth, and append the node's value to result[depth].

def dfs(node, depth):
    if not node: return
    if depth == len(result): result.append([])
    result[depth].append(node.val)
    dfs(node.left,  depth + 1)
    dfs(node.right, depth + 1)
dfs(root, 0)

This works and is O(n) time, but DFS isn't the natural fit — you have to manually track depth. BFS processes nodes level by level by design, so the grouping comes for free.

Spotting the pattern

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

How do you know when one level ends and the next begins while doing BFS?

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.