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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.node
For example:git
Given n = 5
and edges = [[0, 1], [0, 2], [0, 3], [1, 4]]
, return true
.github
Given n = 5
and edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]
, return false
.微信
Hint:app
n = 5
and edges = [[0, 1], [1, 2], [3, 4]]
, what should your return? Is this case a valid 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
.函数
给定n个标记为0到n-1的节点和一个无向边列表(每一个边都是一对节点),编写一个函数来检查这些边是否构成有效的树。this
例如:spa
若是n=5且边数为[[0,1],[0,2],[0,3],[1,4]],则返回true。code
若是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链接。换句话说,任何没有简单循环的连通图都是一棵树。”
注意:能够假定边中不会出现重复的边。由于全部边都是无向的,[0,1]与[1,0]相同,所以不会在边中一块儿出现。
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 }