#429

Medium

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

N-ary Tree Level Order Traversal

Perform a level order traversal on an n-ary tree, capturing nodes by depth using breadth-first search with careful state tracking.

tree breadth first search

Problem Statement

Given an n-ary tree, return a list of lists representing its nodes' values in level order from top to bottom. Each inner list should contain all nodes at a specific depth, preserving left-to-right ordering within that level.

The tree input is serialized in level order, with null markers separating each group of children. For example, given root = [1,null,3,2,4,null,5,6], the expected output is [[1],[3,2,4],[5,6]]. Constraints include a tree height of at most 1000 and a total node count between 0 and 10^4.

Examples

Example 1

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

Output: [[1],[3,2,4],[5,6]]

Example details omitted.

Example 2

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]

Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

Example details omitted.

Constraints

The height of the n-ary tree is less than or equal to 1000
The total number of nodes is between [0, 104]

Solution Approach

Breadth-First Search with Queue

Use a queue to traverse nodes level by level. Enqueue the root and then iterate, dequeueing nodes, appending their values to the current level list, and enqueueing all their children. Continue until the queue is empty.

Track Levels Explicitly

Maintain a loop over the queue size at each level to separate nodes accurately. This ensures each inner list corresponds to a single depth, avoiding mixing nodes from different levels.

Handle Null Separators Properly

Pay attention to null markers in input serialization to reconstruct the children groups correctly. Misinterpreting these markers can result in skipped or merged levels.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(N) because each node is visited once. Space complexity is O(W) where W is the maximum width of the tree, determined by the largest level, due to queue storage.

What Interviewers Usually Probe

Expect discussion of BFS and level tracking.
Interviewer may ask about queue versus recursive approaches.
Clarify how you handle multiple children and null separators.

Common Pitfalls or Variants

Common pitfalls

Merging nodes from different levels due to incorrect queue sizing.
Ignoring null markers in the serialized input, causing misaligned levels.
Failing to initialize lists for each level before adding node values.

Follow-up variants

Recursive DFS approach using depth parameter to append to level lists.
Reverse level order traversal for bottom-up grouping.
Level order traversal with alternating left-right or zigzag patterns.

FAQ

What is the most efficient approach for N-ary Tree Level Order Traversal?

Breadth-first search with explicit level tracking ensures O(N) time and handles multiple children correctly.

How should I handle null separators in the input?

Treat null markers as level boundaries to reconstruct children lists without mixing levels.

Can this problem be solved recursively?

Yes, a DFS approach with a depth parameter can append node values to the corresponding level lists.

What are common mistakes when solving this problem?

Mixing nodes from different levels, ignoring null markers, and not initializing level lists are frequent errors.

How does GhostInterview improve solving level order traversal problems?

It guides the BFS setup, ensures level separation, and alerts you to serialization pitfalls unique to n-ary trees.

terminal

Solution

Solution 1: BFS

First, we check if the root node is null. If it is, we return an empty list directly.

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"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""


class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = []
            for _ in range(len(q)):
                root = q.popleft()
                t.append(root.val)
                q.extend(root.children)
            ans.append(t)
        return ans

Solution 2: DFS

We can use the Depth-First Search method to traverse the entire tree.

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"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""


class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = []
            for _ in range(len(q)):
                root = q.popleft()
                t.append(root.val)
                q.extend(root.children)
            ans.append(t)
        return ans

Continue Practicing

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