LeetCode 261. 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;
    }
}