LeetCode | 0559. N 叉树的最大深度【Python】
Problem
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.
Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).
Example 1:
Input: root = [1,null,3,2,4,null,5,6] Output: 3
Example 2:
Input: root = [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: 5
Constraints:
- The depth of the n-ary tree is less than or equal to
1000. - The total number of nodes is between
[0, 104].
问题
给定一个 N 叉树，找到其最大深度。
最大深度是指从根节点到最远叶子节点的最长路径上的节点总数。
N 叉树输入按层序遍历序列化表示，每组子节点由空值分隔（请参见示例）。
示例 1：
输入：root = [1,null,3,2,4,null,5,6] 输出：3
示例 2：
输入：root = [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] 输出：5
提示：
- 树的深度不会超过 1000 。
- 树的节点数目位于 [0, 10^4] 之间。
思路
DFS
Python3 代码
"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution:
def maxDepth(self, root: 'Node') -> int:
# DFS
if not root:
return 0
depth = 0
for child in root.children:
depth = max(self.maxDepth(child), depth)
return depth + 1 BFS
Python3 代码
"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution:
def maxDepth(self, root: 'Node') -> int:
# BFS
import collections
# 特判，不写会报错
if not root:
return 0
q = collections.deque()
q.append(root)
depth = 0
while q:
for _ in range(len(q)):
node = q.popleft()
for child in node.children:
q.append(child)
depth += 1
return depth GitHub 链接
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