How do you solve N-ary Tree Level Order Traversal on LeetCode?

N-ary Tree Level Order Traversal (LeetCode #429) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Breadth-first search is ideal for level order traversal.

What is the brute force solution for N-ary Tree Level Order Traversal?

The brute force approach for N-ary Tree Level Order Traversal is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we can traverse the tree level by level using a simple recursive function. We collect nodes at each level and store them in a list. This approach is straightforward but inefficient for larger trees. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve N-ary Tree Level Order Traversal?

N-ary Tree Level Order Traversal uses the following concepts: Tree, Breadth-First Search. The recommended patterns to study are: Breadth-First Search, Queue.

What are the interview tips for N-ary Tree Level Order Traversal?

Explain your thought process clearly. Consider edge cases like an empty tree. Practice writing code on a whiteboard or in an online editor.

What are common mistakes when solving N-ary Tree Level Order Traversal?

Forgetting to check if the root is null. Not handling the children correctly in the queue.

#429

N-ary Tree Level Order Traversal

Medium

TreeBreadth-First SearchBreadth-First SearchQueue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a queue to perform a breadth-first traversal of the tree. This allows us to efficiently process nodes level by level, ensuring we capture all nodes at each level before moving to the next.

⚙️

Algorithm

4 steps

1Step 1: Initialize a queue and add the root node to it.
2Step 2: While the queue is not empty, determine the number of nodes at the current level.
3Step 3: Dequeue each node, add its value to the current level list, and enqueue all its children.
4Step 4: After processing all nodes at the current level, add the level list to the result.

solution.py23 lines

1# Full working Python code
2from collections import deque
3
4class Node:
5    def __init__(self, val=None, children=None):
6        self.val = val
7        self.children = children if children is not None else []
8
9class Solution:
10    def levelOrder(self, root):
11        if not root:
12            return []
13        result = []
14        queue = deque([root])
15        while queue:
16            level_size = len(queue)
17            level_nodes = []
18            for _ in range(level_size):
19                node = queue.popleft()
20                level_nodes.append(node.val)
21                queue.extend(node.children)
22            result.append(level_nodes)
23        return result

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the queue storing nodes at the current level.

1Breadth-first search is ideal for level order traversal.
2Using a queue ensures we process nodes level by level.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.