LeetCode 429: N-ary Tree Level Order Traversal Solution in Python – A Step-by-Step Guide

What Is LeetCode 429: N-ary Tree Level Order Traversal?

In LeetCode 429: N-ary Tree Level Order Traversal, you’re given the root of an N-ary tree, where each Node has a value (val) and a list of children (children), and you need to return a list of lists containing the node values in level order. Level order means visiting all nodes at each depth (or level) from top to bottom, left to right within each level. For example, a tree with root 1, children [3, 2, 4], and 3’s children [5, 6] yields [[1], [3, 2, 4], [5, 6]]. Unlike binary trees with fixed left and right kids, N-ary trees allow any number of children per node, making traversal a bit more flexible.

Problem Statement

• Input: A Node object (root of an N-ary tree).
• Output: A list of lists of integers—node values by level, top to bottom, left to right.
• Rules:
• Traverse level by level (breadth-first order).
• Each node may have 0 or more children.
• Return empty list if tree is empty.

Constraints

• Number of nodes is in range [0, 10⁴].
• 0 <= Node.val <= 10⁴.
• Tree height ≤ 1000.

Examples

• Input: [1,null,3,2,4,null,5,6]
• Output: [[1],[3,2,4],[5,6]].
• Input: [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]].
• Input: []
• Output: [].

Understanding the Problem: Traversing Level by Level

To solve LeetCode 429: N-ary Tree Level Order Traversal in Python, we need to visit all nodes in an N-ary tree level by level, collecting their values in a list of lists, where each sublist represents a level from top to bottom, left to right. A naive idea might be to visit nodes randomly and sort by depth—but level order requires a systematic approach! We’ll use:

• Best Solution (BFS with Queue): O(n) time, O(w) space (w = max width)—processes levels efficiently.
• Alternative Solution (DFS with Recursion): O(n) time, O(h) space (h = height)—builds depth-first.

Let’s dive into the BFS solution with a clear, step-by-step explanation.

Best Solution: BFS with Queue

Why This Is the Best Solution

The BFS (Breadth-First Search) with queue method is the top pick because it’s efficient—O(n) time, O(w) space (where w is the maximum width of the tree)—and naturally aligns with level-order traversal by processing all nodes at each level before moving deeper. It uses a queue to manage nodes level-by-level, collecting values as it goes, making it intuitive and optimal for this task. It’s like sweeping through the tree floor-by-floor, gathering everyone at each stop!

How It Works

Think of the tree as a multi-story building you’re exploring:

• Step 1: Initialize:
• Start with a queue containing the root (if exists).
• Prepare a result list to store level values.
• Step 2: Process Levels:
• While queue isn’t empty:
• Get current level size (nodes at this depth).
• Dequeue each node, add its value to a level list.
• Enqueue all its children.
• Append level list to result.
• Step 3: Return Result:
• Return the list of level lists.
• Step 4: Why This Works:
• Queue ensures breadth-first order (level-by-level).
• Level size separates depths cleanly.
• It’s like a guided tour through the tree’s floors!

Step-by-Step Example

Example: [1,null,3,2,4,null,5,6]

• Tree:

1 /|\ 3 2 4 / \ 5 6

• BFS:
• Queue: [1], result = [].
• Level 0: Dequeue 1, val = [1], enqueue [3, 2, 4], result = [[1]].
• Queue: [3, 2, 4].
• Level 1: Dequeue 3 (val = [3], enqueue [5, 6]), 2 (val = [3, 2]), 4 (val = [3, 2, 4]), result = [[1], [3, 2, 4]].
• Queue: [5, 6].
• Level 2: Dequeue 5 (val = [5]), 6 (val = [5, 6]), result = [[1], [3, 2, 4], [5, 6]].
• Queue: [].
• Result: [[1], [3, 2, 4], [5, 6]].

Example: [1,null,2,3,4]

• BFS:
• Queue: [1], result = [[1]].
• Queue: [2, 3, 4], result = [[1], [2, 3, 4]].
• Queue: [], done.
• Result: [[1], [2, 3, 4]].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down so you can follow every step:

# Definition for a Node.
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: 'Node') -> List[List[int]]:
        if not root:
            return []

        # Step 1: Initialize queue and result
        from collections import deque
        queue = deque([root])
        result = []

        # Step 2: Process levels with BFS
        while queue:
            level_size = len(queue)
            level_vals = []

            # Process all nodes at current level
            for _ in range(level_size):
                node = queue.popleft()
                level_vals.append(node.val)
                # Add all children to queue
                for child in node.children:
                    queue.append(child)

            # Add current level to result
            result.append(level_vals)

        # Step 3: Return result
        return result

• Line 2-6: Node class (given):
• val: Node value.
• children: List of child nodes (default empty).
• Line 10-12: Handle empty tree, return [].
• Line 15-17: Initialize:
• queue: Deque with root (e.g., [1]).
• result: List for level values.
• Line 20-30: BFS loop:
• Line 21-22: Get current level size (e.g., 1 for root).
• Line 23: Temp list for level values.
• Line 25-29: Process level:
• Dequeue node, add val (e.g., 1).
• Enqueue all children (e.g., [3, 2, 4]).
• Line 31: Append level to result (e.g., [1]).
• Line 34: Return final list.
• Time Complexity: O(n)—each node processed once.
• Space Complexity: O(w)—queue holds max width (w).

This is like sweeping through a tree’s floors with a queue-guided broom!

Alternative Solution: DFS with Recursion

Why an Alternative Approach?

The DFS (Depth-First Search) with recursion method traverses the tree depth-first, tracking the level of each node and building the result list incrementally. It’s O(n) time and O(h) space (h = height)—intuitive but less aligned with level-order’s breadth-first nature. It’s like climbing down each branch, noting levels as you go, then sorting them out later!

How It Works

Think of it as diving deep:

• Step 1: Recurse with level tracking.
• Step 2: Append values to a level-indexed list.
• Step 3: Return the collected levels.

Step-by-Step Example

Example: [1,null,3,2,4,null,5,6]

• DFS:
• Start: level = 0, result = [[]].
• Node 1: result = [[1]].
• Recurse 3, 2, 4 (level 1): result = [[1], [3, 2, 4]].
• Recurse 3’s kids 5, 6 (level 2): result = [[1], [3, 2, 4], [5, 6]].
• Result: [[1], [3, 2, 4], [5, 6]].

Code for DFS Approach

class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        if not root:
            return []

        # Step 1: Initialize result
        result = []

        # Step 2: DFS with level tracking
        def dfs(node, level):
            if not node:
                return
            # Extend result if new level
            if level == len(result):
                result.append([])
            # Add value to current level
            result[level].append(node.val)
            # Recurse on all children
            for child in node.children:
                dfs(child, level + 1)

        dfs(root, 0)
        return result

• Time Complexity: O(n)—each node visited once.
• Space Complexity: O(h)—recursion stack.

It’s a depth-first level-builder!

Comparing the Two Solutions

• BFS with Queue (Best):
• Pros: O(n), O(w), natural level order, efficient.
• Cons: Queue overhead.
• DFS with Recursion (Alternative):
• Pros: O(n), O(h), simple recursion.
• Cons: Less intuitive for level order.

BFS wins for alignment and clarity.

Additional Examples and Edge Cases

• []: [].
• [1]: [[1]].
• [1,null,2]: [[1], [2]].

BFS handles all.

Complexity Breakdown

• BFS: Time O(n), Space O(w).
• DFS: Time O(n), Space O(h).

BFS is optimal for level order.

Key Takeaways

• BFS: Sweep the levels!
• DFS: Dive and track!
• N-ary: Kids abound.
• Python Tip: Queues rock—see [Python Basics](/python/basics).

Final Thoughts: Traverse That Tree

LeetCode 429: N-ary Tree Level Order Traversal in Python is a level-sweeping adventure. BFS with queue nails it fast, while DFS with recursion dives it deep. Want more tree fun? Try LeetCode 102: Binary Tree Level Order Traversal or LeetCode 589: N-ary Tree Preorder Traversal. Ready to traverse? Head to Solve LeetCode 429 on LeetCode and sweep those levels today!