Leetcode 559. Maximum Depth of N-ary Tree

DFS & BFS 풀이방식

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

# Runtime: 40 ms, faster than 90.09% of Python3 online submissions for Maximum Depth of N-ary Tree.
# Memory Usage: 15.8 MB, less than 100.00% of Python3 online submissions for Maximum Depth of N-ary Tree.
# class Solution:
#     def maxDepth(self, root: 'Node') -> int:
#         if not root: return 0
#         if not root.children: return 1
#         self.maxDepth = 1
#         depth = 1

#         def dfs(node, depth):
#             if not node or not node.children:
#                 self.maxDepth = max(self.maxDepth, depth)
#                 depth = 0
#                 return
#             depth += 1
            
#             if node.children:
#                 for child in node.children:
#                     dfs(child, depth)
        
#         dfs(root, depth)
        
#         return self.maxDepth
    
class Solution:
    def maxDepth(self, root: 'Node') -> int:
        if not root: return 0
        q = [root, None]
        depth = 0
        
        while q:
            node = q.pop(0)
            if node is None:
                depth += 1
                if q: q.append(None)
                continue
                
            if node and node.children:
                for child in node.children:
                    q.append(child)
        
        return depth

https://leetcode.com/problems/maximum-depth-of-n-ary-tree/submissions/