原题链接在这里：https://leetcode.com/problems/graph-valid-tree/
题目：
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.
题解：
Union-Find, 与Number of Islands II相似.
若是数先判断 边的数量是不是 n-1.
然后判断有没有环，若是find(edge[0],edge[1])返回true 说明edge[0], edge[1]两个点之前就连在一起了.
Time Complexity: O(n*logn). Space: O(n).
AC Java:
1 public class Solution { 2 int [] parent; 3 int [] size; 4 5 public boolean validTree(int n, int[][] edges) { 6 if(n < 0 || edges == null || edges.length != n-1){ 7 return false; 8 } 9 parent = new int[n]; 10 size = new int[n]; 11 for(int i = 0; i<n ;i++){ 12 parent[i] = i; 13 size[i] = 1; 14 } 15 16 for(int [] edge : edges){ 17 if(!find(edge[0], edge[1])){ 18 union(edge[0], edge[1]); 19 }else{ 20 return false; 21 } 22 } 23 return true; 24 } 25 26 private boolean find(int i, int j){ 27 return root(i) == root(j); 28 } 29 private int root(int i){ 30 while(i != parent[i]){ 31 parent[i] = parent[parent[i]]; 32 i = parent[i]; 33 } 34 return i; 35 } 36 private void union(int p, int q){ 37 int i = root(p); 38 int j = root(q); 39 if(size[i] < size[j]){ 40 parent[i] = j; 41 size[j] += size[i]; 42 }else{ 43 parent[j] = i; 44 size[i] += size[j]; 45 } 46 } 47 }
原文：http://www.cnblogs.com/Dylan-Java-NYC/p/5237914.html