N-ary Tree Level Order Traversal — LeetCode 429 BFS Pattern

Problem Statement#

Given the root of an n-ary tree, return the level order traversal of its nodes' values. Each level should appear as its own sub-array.

Constraints:

• The total number of nodes is in the range [0, 10^4]
• The depth of the n-ary tree is at most 1000

Input:  root = [1,null,3,2,4,null,5,6]
Output: [[1],[3,2,4],[5,6]]

Input:  root = []
Output: []

Why This Problem Matters#

LeetCode 429 — N-ary Tree Level Order Traversal — is a staple BFS warmup at Amazon, Meta, Microsoft, and Bloomberg. It generalizes LC 102 Binary Tree Level Order Traversal to nodes with arbitrary numbers of children — a setting that mirrors real-world hierarchies like file systems, organizational charts, and DOM trees.

Interviewers use this problem to confirm that candidates can handle BFS with grouped levels, manage queue size correctly, and adapt to non-binary structures. It is often the first round of an Amazon SDE-1 phone screen because it tests basic queue mechanics without requiring tree-DP gymnastics.

The same pattern underpins workflow engines, social network feed expansion, and graph-layer analysis at FAANG-scale infrastructure teams.

The Core Insight#

Standard BFS plus a "level size snapshot" trick: at the start of each iteration, record the queue length. Pop exactly that many nodes (one full level), collect their values into a sub-array, and enqueue all their children. Repeat until the queue is empty.

The only difference from binary BFS is that we iterate node.children instead of node.left, node.right. Everything else — queue, level grouping, output shape — is identical.

Visual Dry Run#

Tree where 1 has children [3, 2, 4] and 3 has children [5, 6]:

Final output: [[1], [3, 2, 4], [5, 6]].

Solution (Optimal)#

from collections import deque
from typing import List, Optional
 
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children if children is not None else []
 
class Solution:
    def levelOrder(self, root: Optional[Node]) -> List[List[int]]:
        if root is None:
            return []
        result: List[List[int]] = []
        queue = deque([root])
        while queue:
            level_size = len(queue)
            level_vals: List[int] = []
            for _ in range(level_size):
                node = queue.popleft()
                level_vals.append(node.val)
                for child in node.children:
                    queue.append(child)
            result.append(level_vals)
        return result

var levelOrder = function(root) {
    if (!root) return [];
    const result = [];
    const queue = [root];
    while (queue.length > 0) {
        const levelSize = queue.length;
        const levelVals = [];
        for (let i = 0; i < levelSize; i++) {
            const node = queue.shift();
            levelVals.push(node.val);
            for (const child of node.children) {
                queue.push(child);
            }
        }
        result.push(levelVals);
    }
    return result;
};

Time: O(n) — every node is enqueued and dequeued exactly once. Space: O(n) — queue can hold up to one full level (worst case the leaf level).

Common Mistakes#

• Forgetting the level-size snapshot — without it, results merge into one flat list.
• Using queue.shift() in JavaScript on huge inputs is O(n); use a deque-style index pointer for performance-critical scenarios.
• Returning [[]] instead of [] when root is null.
• Iterating children with index when the children array could contain None placeholders from a serialized form (only present in the test harness, not in node.children).
• Recursive DFS with manual depth tracking works but is harder to reason about; BFS is the canonical answer.

Interview Tips#

• Sketch the queue and pointer to the level break.
• Mention the alternative DFS solution that passes a depth index and appends to result[depth]. Both are O(n).
• Emphasize that this generalizes to k-ary trees, file system traversal, and BFS over any DAG.
• If asked, code BFS first, then DFS to show range.

Follow-up Questions#

• "Bottom-up level order" — Reverse the result list (LC 107 style).
• "Average per level" — LC 637, replace level_vals with running sum and divide.
• "Right-side view of n-ary tree" — Push only the last node of each level.
• "Zigzag traversal" — Alternate appending vs. prepending per level (LC 103 generalized).
• "Memory-bounded BFS for very wide trees" — Stream level by level to disk or apply a per-level pruning rule.

Key Takeaways#

• LeetCode 429 N-ary Tree Level Order Traversal is solved with classic BFS plus a level-size snapshot at each iteration.
• Time is O(n), space is O(n) for the queue at the widest level.
• The only generalization from binary BFS is iterating node.children instead of left and right.
• A DFS solution with explicit depth indexing also works and is O(n).
• Common follow-ups include zigzag, right-side view, and per-level averages.
• This is a frequent Amazon SDE-1 phone screen and Meta E3 warmup question.
• The level-snapshot trick generalizes to BFS over DAGs, file systems, and DOM trees.