LeetCode 1522 Diameter of N-Ary Tree

LeetCode

Given a root of an N-ary tree, you need to compute the length of the diameter of the tree.

The diameter of an N-ary tree is the length of the longest path between any two nodes in the tree. This path may or may not pass through the root.

(Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value.)

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: 3
Explanation: Diameter is shown in red color.

Example 2:

Input: root = [1,null,2,null,3,4,null,5,null,6]
Output: 4

Example 3:

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: 7

Constraints:

The depth of the n-ary tree is less than or equal to 1000. The total number of nodes is between [1, 10<sup>4</sup>].

题目大意：

求树的直径：任何两个节点的最大距离

解题思路：

类似于LeetCode 543 Diameter of Binary Tree，但此题为N叉树

解题步骤：

DFS

注意事项：

1. 求数组中最大的两数和，用去掉最大值的方法得到次大值。还要注意初始值加入[1, 1]，避免没有儿子节点或只有一个的情况

Python代码：

def diameter(self, root: 'Node') -> int:
	max_len = 0

	def dfs(root):
		if not root:
			return 0
		nonlocal max_len
		path_len = [1, 1]
		for child in root.children:
			path_len.append(dfs(child) + 1)
		largest = max(path_len)
		path_len.remove(largest)
		second_largest = max(path_len)
		total = largest + second_largest - 1
		max_len = max(total, max_len)
		return largest

	dfs(root)
	return max_len - 1

算法分析：

时间复杂度为O(n)，空间复杂度O(n)