Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.
For example:
Given n = 5 and edges = [[0, 1], [0, 2], [0, 3], [1, 4]], return true.
Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]], return false.
Hint:
- Given
n = 5andedges = [[0, 1], [1, 2], [3, 4]], what should your return? Is this case a valid tree? - According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”
Note: you can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.
给定n个标记为0到n-1的节点和一个无向边列表（每个边都是一对节点），编写一个函数来检查这些边是否构成有效的树。
例如：
如果n=5且边数为[[0，1]，[0，2]，[0，3]，[1，4]]，则返回true。
如果n=5且边数为[[0，1]，[1，2]，[2，3]，[1，3]，[1，4]]，则返回false。
提示：
考虑到n=5和edges=[[0，1]，[1，2]，[3，4]]，你应该返回什么？这是一个有效的树吗？
根据维基百科上树的定义：“树是一个无向图，其中任何两个顶点都由一条路径连接。换句话说，任何没有简单循环的连通图都是一棵树。”
注意：可以假定边中不会出现重复的边。因为所有边都是无向的，[0，1]与[1，0]相同，因此不会在边中一起出现。
BFS
1 class Solution {
2 func validTree(_ n:Int,_ edges:[[Int]]) -> Bool{
3 var g:[Set<Int>] = [Set<Int>](repeating:Set<Int>(),count:n)
4 var s:Set<Int> = [0]
5 var q:[Int] = [0]
6 for a in edges
7 {
8 g[a[0]].insert(a[1]);
9 g[a[1]].insert(a[0]);
10 }
11 while(!q.isEmpty)
12 {
13 var t:Int = q.first!
14 q.removeFirst()
15 for a in g[t]
16 {
17 if s.contains(a)
18 {
19 return false
20 }
21 s.insert(a)
22 q.append(a)
23 g[a].remove(t)
24 }
25 }
26 return s.count == n
27 }
28 }
原文地址：https://www.cnblogs.com/strengthen/p/10227256.html