Graph Valid Tree

犯错：这个题一开始不用check edge的合法性，因为n=1的时候，没有edge是true的，n=2时没有edge是不合法的，所以直接让n和edge的关系去判断就好了

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.

Example 1:

Input: n = 5, and edges = [[0,1], [0,2], [0,3], [1,4]]
Output: true

Example 2:

Input: n = 5, and edges = [[0,1], [1,2], [2,3], [1,3], [1,4]]
Output: false

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.

class Solution { public boolean validTree(int n, int[][] edges) { if(n <= 0){ return false; }

valid tree: edge 数比node数 小1， 无环， 用bfs判断，结束以后，bfs遍历的点，一定和node数一样才对

    if(edges.length != n-1)
        return false;


    Map<Integer,Set<Integer>> map = buildTree(n,edges);

    Queue<Integer> queue = new LinkedList<>();

    Set<Integer> hash = new HashSet<>();

    queue.offer(0);
    hash.add(0);

    while(!queue.isEmpty()){
        int node = queue.poll();

        for(int nb : map.get(node)){
            if(hash.contains(nb)){
                continue;
            }

            hash.add(nb);
            queue.offer(nb);
        }
    }

    return hash.size() == n;
}

public Map<Integer,Set<Integer>> buildTree(int n, int[][] edges){
    Map<Integer,Set<Integer>> map = new HashMap<>();

    for(int i = 0; i < n;i++){
        map.put(i, new HashSet<>());
    }

    for(int i = 0; i < edges.length;i++){
        int u = edges[i][0];
        int v = edges[i][1];

        map.get(u).add(v);
        map.get(v).add(u);
    }

    return map;
}

}