p20 — Binary Tree Level Order Traversal

Source: LeetCode 102 · Medium · Topics: Tree, BFS, Queue Companies (2024–2025 frequency): Amazon (very high), Meta (very high), Google (high), Microsoft (high), Bloomberg (high), Apple (medium), LinkedIn (medium), TikTok/ByteDance (medium) Loop position: very common phone screen or first round; opens an entire family — LC 103 (zigzag), 107 (bottom-up), 199 (right side view), 314 (vertical order), 116 (next right pointers).

1. Quick Context

This is the canonical BFS template for trees. Output is a list of lists where each inner list contains the values at one depth level, top to bottom, left to right.

The whole problem hinges on ONE trick: freeze the queue’s size before processing each level. Without that, BFS works but you can’t distinguish where one level ends and the next begins.

while queue:
    n = len(queue)           # ← snapshot
    level = []
    for _ in range(n):       # ← drain exactly n nodes; the queue may grow during this loop
        node = queue.popleft()
        level.append(node.val)
        if node.left:  queue.append(node.left)
        if node.right: queue.append(node.right)
    result.append(level)

That snapshot pattern is the entire problem. Once you have it, you have LC 103, 107, 199, 116, 117 — they’re all variations on what you do with the level.

What it looks like it tests: can you do BFS. What it actually tests: do you internalize the “snapshot the queue size before draining” pattern, which appears in every level-aware graph BFS (Word Ladder, Rotting Oranges, Open the Lock, Bus Routes, Shortest Path in Binary Matrix).

2. LeetCode Link + Attempt Gate

🔗 https://leetcode.com/problems/binary-tree-level-order-traversal/

STOP. 10-min timer. Don’t peek at hints; this is the gateway problem to every BFS-flavored question.

3. Prerequisite Concepts

• p15 — recursive tree traversal (we contrast against it here)
• phase-01-foundations/labs/lab-05-stack-queue-applications.md — collections.deque as O(1)-pop queue
• phase-01-foundations/labs/lab-09-tree-traversal-fundamentals.md — why BFS vs DFS

4. How to Approach (FRAMEWORK Steps 1–9)

Step 1 — Restate: “Given the root of a binary tree, return its node values level by level, top to bottom, left to right, as a list of lists.”

Step 2 — Clarify:

• “Empty tree?” (Return [] — not [[]].)
• “Single node?” (Return [[root.val]].)
• “Within a level, ordering left-to-right matters?” (Yes — LC enforces this.)
• “Should the output type be List[List[int]] exactly?” (Yes.)

Step 3 — Constraints: Up to 2000 nodes. O(N) time, O(W) space where W = maximum width of the tree (could be N/2 for a complete tree).

Step 4 — Examples:

• [3,9,20,null,null,15,7] → [[3],[9,20],[15,7]].
• [1] → [[1]].
• [] → [].
• Left-skewed 1→2→3 → [[1],[2],[3]].
• Right-skewed 1→2→3 → [[1],[2],[3]].

Step 5 — Brute Force: DFS with depth tracking — recurse with (node, depth), append node.val into result[depth] (allocate a new sublist when depth == len(result)). This is genuinely O(N) and works. Mention it; ship the BFS as primary because it matches the problem’s intent and generalizes cleanly to graph problems.

Step 6 — Complexity: O(N) time (each node enqueued and dequeued once), O(W) space for the queue at the widest level (worst case W = N/2 for a complete tree).

Step 7 — Pattern Recognition: “Process layer by layer” → BFS with level-size snapshot. This is the gateway pattern. It scales unchanged to graph BFS, multi-source BFS, and bidirectional BFS.

Step 8 — Develop:

from collections import deque

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

Step 9 — Test: Empty, single, balanced canonical example, left-skewed, right-skewed.

5. Progressive Hints

If stuck for 5 min, open hints.md.

6. Deeper Insight — Why the Snapshot Matters

If you omit n = len(q) and just drain until empty, you can still recover the level structure by attaching a depth field to every queued node (q.append((child, depth + 1))) and bucketing the results by depth. That works but pays an O(N) memory cost for the depths AND requires you to bucket after the fact.

The snapshot trick is structurally cleaner: it uses the queue’s instantaneous size as an implicit level marker. The BFS frontier always contains exactly the nodes at the current depth (plus possibly some from the next, mid-drain). Snapshotting captures “exactly this layer” without per-node bookkeeping.

This generalizes to:

• LC 127 Word Ladder — same snapshot to count BFS distance (the answer is the level index where we first hit endWord).
• LC 994 Rotting Oranges — multi-source BFS where each “level” = one minute.
• LC 752 Open the Lock — BFS over combination space; answer = level we first reach target.
• LC 433 Minimum Genetic Mutation — same shape as Word Ladder.

The “level = step count” interpretation is what makes BFS the shortest-path algorithm for unweighted graphs. This problem teaches you that pattern.

7. Anti-Pattern Analysis

1. Using a list as a queue. list.pop(0) is O(N). Use collections.deque — popleft() is O(1). Junior-level mistake; named at interview.
2. DFS with no depth tracking. If you DFS without recording depth, you can’t group by level. The fix is dfs(node, depth) writing into result[depth]. Acceptable as alternative; mention it.
3. Mutating len(q) inside the for-loop. for _ in range(len(q)) is correct (n is captured at loop start). But while len(q) > 0 inside the inner loop would mix levels. State the snapshot pattern explicitly.
4. Recursive BFS. “Recursive BFS” is a contradiction in terms — recursion uses an implicit stack (LIFO), BFS needs a queue (FIFO). Anyone claiming “recursive BFS” actually means DFS with depth tracking. Be precise.
5. Returning [[]] for empty input. Read the problem: empty tree → [], not [[]]. Off-by-one error.

8. Skills & Takeaways

• The BFS level-size snapshot pattern — your single most-used template in interview BFS problems.
• collections.deque for O(1) queue operations — never use a list as a queue.
• DFS+depth as the symmetric alternative — gives the same result with implicit-stack memory and trivial to code recursively.
• “BFS level = distance” — the conceptual key to shortest-path algorithms on unweighted graphs.
• Reading output schemas carefully — [] vs [[]] distinguishes a careful candidate from a sloppy one.

9. When to Move On

Move on when:

• You can write the BFS snapshot version from memory in <90 seconds.
• You can write the DFS+depth alternative on a whiteboard without bugs.
• You can do LC 199 (Right Side View — return the rightmost element of each level) as a 30-second variation.
• You can do LC 103 (zigzag) — flip every other level — without restructuring the algorithm.
• You can do LC 107 (bottom-up) by either reversing at the end or pre-pending into the result.
• Your stress test confirms BFS and DFS produce identical level lists on 1000 random binary trees.

10. Company Context

Amazon:

• Asks this verbatim almost every cycle for SDE-I / SDE-II. Bar Raisers often pivot to LC 199 (right side view) or LC 314 (vertical order) as the follow-up.
• Watch for: “now output one string per level, comma-separated” — trivial change, tests whether you panic at format tweaks.
• Scorecard phrase: “Coding Excellence — clean BFS template; correctly distinguished [] from [[]].”

Meta:

• Will demand LC 199 as the immediate follow-up. Some interviewers ask LC 116 (next right pointers) instead — same algorithm, you wire prev.next = curr during the level drain.
• Scorecard phrase: “Strong — recognized snapshot pattern; chained to LC 199/116 without restructuring.”

Google:

• May skip 102 entirely and start with LC 314 (vertical order traversal) which composes BFS-by-level with column tracking. Knowing this problem fluently is the entry ticket.
• Scorecard phrase: “Pattern Recognition — generalized level-BFS to column-indexed BFS for vertical order.”

Microsoft:

• Often paired with LC 103 (zigzag). Asks “can you do it without an explicit reverse?” → answer: alternate appendleft / append on the level list. Shows queue-vs-deque thinking.
• Scorecard phrase: “Algorithm — eliminated unnecessary reverse via deque trick.”

Bloomberg:

• Frames as “render a financial tree (parent positions → child positions) level by level for UI display.” Output format is verbose; tests pragmatism.
• Scorecard phrase: “Pragmatic — clean BFS adapted to UI rendering output.”

Apple:

• Asks “what if the tree is enormous and you want to stream level-by-level instead of materializing the whole result?” → answer: yield each level as a generator (yield level instead of out.append(level)). Memory becomes O(W) instead of O(N).
• Scorecard phrase: “Solid — recognized streaming / generator opportunity.”

LinkedIn:

• Common Q4 / Q5 of phone screen. Asks LC 102 first, LC 107 (bottom-up) second.
• Scorecard phrase: “Standard — solved both 102 and 107 fluently.”

TikTok/ByteDance:

• Goes straight to LC 199 (right side view). Reject if you can’t transcribe the BFS template under pressure.
• Scorecard phrase: “Pass — produced right-side-view at first attempt.”

11. Interviewer’s Lens

12. Level Delta

Mid (L4):

• Writes BFS with deque + snapshot first try.
• Handles empty case correctly.
• Knows LC 199 is “return last of each level”.

Senior (L5):

• Names the snapshot pattern explicitly.
• Sketches DFS+depth as alternative within 30 seconds when asked.
• Fluent in 102 → 199 → 103 progression.

Staff (L6):

• Articulates “BFS level = shortest path in unweighted graph” without prompting.
• Suggests generator-based streaming for memory-bounded environments.
• Walks through LC 116 (next right pointers) using the same template + level-internal wiring.

Principal (L7+):

• Connects to graph BFS family (Word Ladder, Rotting Oranges, Open the Lock).
• Discusses when BFS exceeds practical memory (very wide trees) and the trade-off with iterative deepening DFS.
• Mentions the parallel BFS variant — frontier as a level-by-level synchronization point — and its relevance to GPU graph algorithms.

13. Follow-up Q&A

Q1: Without using BFS, can you solve this with DFS? A: Yes. Recurse with (node, depth); on entry, if depth == len(result), append a new empty list to result; then append node.val to result[depth]. O(N) time, O(H) stack. Produces identical output. Trade-off: deeper trees → more stack; very wide trees → BFS frontier exceeds DFS stack. Choose by tree shape if memory is bounded.

Q2: LC 199 — return only the rightmost element of each level. A: Two ways.

• BFS: same template, but level[-1] is the answer for each level. Or for memory: keep only the last node.val seen in each level-drain.
• DFS: visit right-before-left; track depth; the first node seen at each depth IS the right-side. Single pass, O(H) memory.

Q3: LC 103 (zigzag) — alternate left-to-right and right-to-left per level. A: Track an lr_flag boolean. Same BFS; on odd levels, build the level list using appendleft on a deque (or reverse() at end). Same O(N) / O(W).

Q4: LC 116 — populate next pointers to the right neighbor at each level. A: Modify the level drain: keep prev = None at level start; for each node popped, set prev.next = node (if prev), then prev = node. End of level: prev.next = None (or just leave None as default). Same template; the only addition is in-loop wiring.

Q5: The tree has billions of nodes — you can’t materialize the result. What’s the change? A: Convert level_order to a generator: yield level instead of out.append(level). Caller iterates and processes one level at a time. Memory drops from O(N) to O(W). This is the streaming pattern that unlocks LC-style problems for true big-data scale.

14. Solution Walkthrough

See solution.py:

• level_order_bfs(root) — canonical BFS with deque + level-size snapshot (ship this).
• level_order_dfs(root) — DFS+depth alternative (recursive, indexes into result[depth]).
• right_side_view(root) — LC 199 follow-up via BFS (last-in-each-level).
• zigzag_level_order(root) — LC 103 follow-up via BFS + alternating direction.
• build_tree_from_list(vals) — level-order list → TreeNode (uses is_left: bool toggle pattern — does NOT set attributes on TreeNode because __slots__). Defensive pattern; see p16 for the lesson.
• stress_test() — 1000 random binary trees; BFS == DFS, every time.
• edge_cases() — empty, single, left-skewed, right-skewed, canonical LC example.

Run: python3 solution.py → “ALL TESTS PASSED”.

15. Beyond the Problem

Principal-engineer code review comment:

“The BFS is correct, but a few things. First: capture n = len(q) into a local before the for-loop — not in range(len(q)) re-call — because while range is evaluated once, the explicit local variable signals intent to the reader and reads better in code review. Second: please use collections.deque, not list.pop(0) — I’ve seen this exact code thrash production at 100k nodes. Third: have you considered yielding each level? Half the time we want to stream this output (e.g., paginating UI rendering) and the materialized result is wasted. Fourth: the test helper that builds a tree from a level-order list should NOT set attributes on TreeNode — we use __slots__ for memory; any code that sets .foo on a TreeNode silently breaks under __slots__. Use external state (a bool flag, a dict from parent → child-count). Last: in production this would be parameterized on a Tree protocol, not concrete TreeNode — but for an interview, this is fine. Ship it.”

Connections forward:

• Phase 4 — graph BFS uses this exact template (queue + visited set + snapshot for distance).
• Phase 4 — multi-source BFS (LC 994 Rotting Oranges) is BFS where the initial queue contains all sources.
• Phase 4 — Word Ladder (LC 127) is BFS where each “level” is one mutation.
• Phase 4 — 0-1 BFS uses a deque for two-cost BFS (close cousin of this).
• LC 116/117 (next right pointers) are direct applications of the level-drain template.
• LC 314 / 987 (vertical order traversal) compose this template with column tracking.

This is the gateway problem. Drill it cold.