LeetCode Graph Valid Tree

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.

Check is edges count is equal to n-1(There is only one cluster after merging).

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:

 public class Solution {
     public boolean validTree(int n, int[][] edges) {
         if(edges == null || edges.length != n-1){
             return false;
         }
 
         UnionFind tree = new UnionFind(n);
         for(int [] edge : edges){
             if(!tree.find(edge[0], edge[1])){
                 tree.union(edge[0], edge[1]);
             }else{
                 return false;
             }
         }
         return true;
     }
 }
 
 class UnionFind{
     int count, n;
     int [] size;
     int [] parent;
 
     public UnionFind(int n){
         this.n = n;
         this.count = n;
         size = new int[n];
         parent = new int[n];
         for(int i = 0; i<n; i++){
             parent[i] = i;
             size[i] = 1;
         }
     }
 
     public boolean find(int i, int j){
         return root(i) == root(j);
     }
 
     private int root(int i){
         while(i != parent[i]){
             parent[i] = parent[parent[i]];
             i = parent[i];
         }
         return i;
     }
 
     public void union(int p, int q){
         int i = root(p);
         int j = root(q);
         if(size[i] > size[j]){
             parent[j] = i;
             size[i] += size[j];
         }else{
             parent[i] = j;
             size[j] += size[i];
         }
         this.count--;
     }
 
     public int size(){
         return this.count;
     }
 }