leetcode Graph Valid Tree

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.

Show Hint 

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.

union-find问题的变种，思路和解决union-find的问题几乎是完全一样的，由于树的定义要求图不能出现环路，因此当find出现相同的根时说明出现回路，返回false，另外一棵树只能有一个根，因此可以设置一个计数器判断最后所剩的根即树的个数是否为1：

private int count;private int[] res;public boolean validTree(int n, int[][] edges) {    res=new int[n];    for(int i=0;i<n;i++)        res[i]=i;    count=n;    for(int i=0;i<edges.length;i++){        int left=find(res,edges[i][0]);        int right=find(res,edges[i][1]);        if(left==right) return false;        res[left]=right;        count--;    }    return count==1;}public int find(int[]res,int i){    while(res[i]!=i) i=res[i];    return i;}