public class Solution {/*** @param n an integer* @param edges a list of undirected edges* @return true if it's a valid tree, or false*/public boolean validTree(int n, int[][] edges) {// Write your code here//To find cycle and not connect nodeint nEdges = edges.length;//If total number of the node is n, for the tree(undirection graph), //the number of the edges should be n - 1. //If there are more than n - 1 edges, there are cycles.//If there are less than n - 1 edges, some nodes are not connected.if ( n != (nEdges + 1)) {return false;}//After the nEdge validation, we need to consider the on has cycles, //but also contains un-fully-connected(which only have one or zero edge) nodes.//So we can create a hashset to remeber all the nodes that is connected.//And a Queue is also created to put the node that got connected.//If there are cycles and breaks, //in the end the node which is not fully connected will not be able to add to the queue//In the end, we just need to compare the size of hashset with n. HashMap<Integer, HashSet<Integer>> graph = initalizeGraph(n,edges); HashSet<Integer> cnntNodes = new HashSet<Integer>();Queue<Integer> queue = new LinkedList<Integer>();queue.offer(0);cnntNodes.add(0);while (!queue.isEmpty()) {int curNode = queue.poll();HashSet<Integer> cnntToThisNode = graph.get(curNode);for (Integer i : cnntToThisNode) {if (cnntNodes.contains(i)) {continue;}cnntNodes.add(i);queue.offer(i);}}return (cnntNodes.size() == n);}private HashMap<Integer, HashSet<Integer>> initalizeGraph(int n, int[][] edges) {HashMap<Integer, HashSet<Integer>> graph = new HashMap<>();for (int i = 0; i < n; i++) {graph.put(i, new HashSet<Integer>());}for (int i = 0; i < edges.length; i++) {int u = edges[i][0];int v = edges[i][1];graph.get(u).add(v);graph.get(v).add(u);}return graph;}}