Binary Tree Level Order Traversal: The BFS Template

"Binary Tree Level Order Traversal" is a Medium-level staple in Amazon, Microsoft, Meta loops, and it is a textbook fit for the Tree BFS pattern — one of the nine patterns that cover the bulk of what FAANG coding interviews actually test. Part 1 of 11 in the Tree BFS arc.

The Problem

Given the root of a binary tree, return the level order traversal of its nodes' values — a list per level, left to right. This is the template question: the snapshot loop you write here is reused, nearly unchanged, by most of the BFS family.

Input:  root = [3, 9, 20, None, None, 15, 7]
Output: [[3], [9, 20], [15, 7]]

Recognizing the Tree BFS Pattern

Breadth-first search explores a tree or graph level by level using a queue, and its defining superpower is the level snapshot: freeze the queue length, process exactly that many nodes, and you have processed one complete level. Anything phrased per-level — level lists, level averages, rightmost visible node, shortest path in unweighted structures — is BFS by construction.

Grouped-by-level output is the least ambiguous BFS cue that exists. The subtlety is the snapshot: len(queue) at the top of each round is exactly the size of the current level, because children enqueued during the round belong to the next one.

The Approach

Seed a deque with the root. Each outer iteration captures len(queue), pops exactly that many nodes — collecting values and enqueuing children — and appends the collected list. When the queue drains, every level has been emitted in order.

Use collections.deque, and pop from the left: list.pop(0) is O(n) per pop and converts the whole traversal to O(n²). Interviewers at Meta specifically look for this detail in Python candidates.

Python Solution

class TreeNode:
    def __init__(self, val: int = 0, left: "TreeNode | None" = None,
                 right: "TreeNode | None" = None):
        self.val = val
        self.left = left
        self.right = right


from collections import deque


def level_order(root: TreeNode | None) -> list[list[int]]:
    """Values grouped by level, left to right."""
    if not root:
        return []
    result: list[list[int]] = []
    queue: deque[TreeNode] = deque([root])
    while queue:
        level: list[int] = []
        for _ in range(len(queue)):        # snapshot: this level only
            node = queue.popleft()
            level.append(node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        result.append(level)
    return result

Complexity

• Time: O(n) — every node enters and leaves the queue exactly once
• Space: O(w) — the queue holds at most one level — w is the tree's max width

Interview Tips and Follow-Ups

• Say 'O(w) queue, up to n/2 on a complete tree' — width-aware space analysis separates practiced candidates from template reciters.
• DFS with a depth parameter can also produce level lists — mention it, note BFS is canonical, and move on.
• Every remaining question in this arc is this loop with one line changed — say that at the end and mean it.

If this clicked, continue the Tree BFS arc in the Technical Interview category. One hundred questions, nine patterns, all in Python.