Union-Find

261. 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.
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.

public class Solution {
    public boolean validTree(int n, int[][] edges) {
        // initialize n isolated islands
        int[] nums = new int[n];
        Arrays.fill(nums, -1);
        
        // perform union find
        for (int i = 0; i < edges.length; i++) {
            int x = find(nums, edges[i][0]);
            int y = find(nums, edges[i][1]);
            
            // if two vertices happen to be in the same set
            // then there's a cycle
            if (x == y) return false;
            
            // union
            nums[x] = y;
        }
        
        return edges.length == n - 1;
    }
    
    int find(int nums[], int i) {
        if (nums[i] == -1) return i;
        return find(nums, nums[i]);
    }
}

注意：并查集可以解决父节点，环路等问题，nums里面存的是每个结点的父节点，找的时候一直寻找到底就可以找到最高的父节点。