标签：queue static number 分享 down stat while des tco
Given a n-ary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
For example, given a 3-ary tree:
We should return its max depth, which is 3.
Note:
1000.5000.
DFS:
"""
# Definition for a Node.
class Node(object):
def __init__(self, val, children):
self.val = val
self.children = children
"""
class Solution(object):
def maxDepth(self, root):
"""
:type root: Node
:rtype: int
"""
maxD=0
if root:
for c in root.children:
maxD=max(self.maxDepth(c),maxD)
return maxD+1
return maxD
BFS:
"""
# Definition for a Node.
class Node(object):
def __init__(self, val, children):
self.val = val
self.children = children
"""
class Solution(object):
def maxDepth(self, root):
"""
:type root: Node
:rtype: int
"""
queue=[root]
de=0
if root:
while queue:
de+=1
for i in range(len(queue)):
node=queue.pop(0)
for c in node.children:
queue.append(c)
return de
[LeetCode&Python] Problem 559. Maximum Depth of N-ary Tree
标签：queue static number 分享 down stat while des tco
原文地址：https://www.cnblogs.com/chiyeung/p/9739585.html
