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.

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.

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.

Solution:

Method 1: BFS

If a graph is a valid tree, it satisfies:

1. If and only if the number of edges equals n - 1.

2. All n vertices are connected.

We use BFS to traverse the graph and store all reached vertices to a HashSet.

After traversing, we check if the size of HashSet is n.



Code:

public class Solution {
    public boolean validTree(int n, int[][] edges) {
        if (n == 0) {
            return false;
        }
        if (edges.length != n - 1) {
            return false;
        }
        
        ArrayList<Integer>[] adj = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new ArrayList<Integer>();
        }
        for (int[] edge : edges) {
            adj[edge[0]].add(edge[1]);
            adj[edge[1]].add(edge[0]);
        }
        
        HashSet<Integer> set = new HashSet<>();
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(0);
        set.add(0);
        
        while (!queue.isEmpty()) {
            int v = queue.poll();
            for (int neighbor : adj[v]) {
                if (!set.contains(neighbor)) {
                    set.add(neighbor);
                    queue.offer(neighbor);
                }
            }
        }
        
        return set.size() == n;
    }
}

Method 2: Union Find


Code: