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



Three states: visiting , visited, unexpolored 
   In number :   1.         2              0

If visiting node’s nei is also visiting , then cycle 







class Solution{
  public boolean validTree(int n, int[][] edges){
    List<List<Integer>> list = new ArrayList<>();
    Queue<Integer> queue = new LinkedList<>();
    int[] visited = new int[n]; // default inital value is 0, means unexplored
    
    for(int i = 0; i < n; i++){
      list.add(new ArrayList<>());
    }
    
    for(int i = 0; i < edges.length; i++){
      int firstNode = edges[i][0];
      int secNode = edges[i][1];
      list.get(firstNode).add(secNode);
      list.get(secNode).add(firstNode);
    }
    queue.offer(0);
    while(!queue.isEmpty()){
      int cur = queue.poll();
      visited[cur] = 1; // means visiting 
      for(Integer nei : list.get(cur)){
        if(visited[nei] == 1){
          return false;
        }
          
        if(visited[nei] == 0){
          visited[nei] = 1;
          queue.offer(nei);
        }
        
        
      }
      visited[cur] = 2; // means visited
    }
    
    // another check for valid tree , is it has only one connected component
    // if no cycle, after all the nodes should be visited, which is 2 in the int[] visited
    // if there's still nodes left that is not 2, or, in other words, if its 0, 
    // then we know we have more than one connected components 
    
    for(int i = 0; i < n; i++){
      if(visited[i] == 0){
        return false;
      }
    }
    
    return true;
  }
}