algoadvance

Given an n-ary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).

Clarifying Questions

To ensure a complete understanding of the problem, I might ask the following questions:

Strategy

To solve this problem, we’ll need to perform a level-order traversal (breadth-first traversal) of the n-ary tree. Here is the step-by-step strategy:

  1. Initialization: Use a queue to facilitate level order traversal. Enqueue the root node along with the level information.
  2. Traversal: Dequeue nodes level by level, recording their values, and enqueue their children for the next level.
  3. Constructing Result: Use a list of lists to collect nodes level by level.
  4. Edge Cases: Handle the scenario where the input tree is empty by returning an empty list.

Code

from typing import List
from collections import deque

# 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 []
        
        result = []
        queue = deque([(root, 0)])  # Queue will store tuples of (node, level)
        
        while queue:
            node, level = queue.popleft()
            
            # If the level is not yet in result, add it
            if level == len(result):
                result.append([])
                
            # Append the current node's value to the corresponding level
            result[level].append(node.val)
            
            # Enqueue all the children with the level + 1
            for child in node.children:
                queue.append((child, level + 1))
        
        return result

Time Complexity

The time complexity of this approach is (O(N)), where (N) is the number of nodes in the tree. Each node is visited exactly once.

Space Complexity

The space complexity is also (O(N)) because, in the worst case, the maximum number of nodes that can be held in the queue at once is proportional to the total number of nodes in the tree, assuming it’s a complete n-ary tree. Additionally, the output list will hold all the node values.

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