261.Graph Valid Tree

  
    /*
     * 261.Graph Valid Tree
     * 2016-3-29 by Buttercola
     * 和上面的BFS方法一样，不过这里是无向图，所以需要两个点都要存对方
     * 图要形成树必须有两个方面，一个是没有cycle另一个是没有孤立的点
     */
     public boolean validTree(int n, int[][] edges) {
            // build the graph using adjacent list
           HashMap<Integer,HashSet<Integer>> graph=new HashMap<Integer,HashSet<Integer>>();
            for(int i = 0; i < n; i++)
                graph.put(i,new HashSet<Integer>());
            for(int[] edge : edges) {
                graph.get(edge[0]).add(edge[1]);
                graph.get(edge[1]).add(edge[0]);
            }
            // no cycle
            boolean[] visited = new boolean[n];
            Queue<Integer> queue = new LinkedList<Integer>();
            queue.add(0);
            while(!queue.isEmpty()) {
                int node = queue.poll();
                if(visited[node])
                    return false;
                visited[node] = true;//保证每个点都访问到
                for(int neighbor : graph.get(node)) {
                    queue.offer(neighbor);//找到所有的邻接点，并且砍掉回来的路
                    graph.get(neighbor).remove((Integer)node);
                }
            }
            // fully connected
            for(boolean result : visited){
                if(!result)
                    return false;
            }
            return true;
        }

 当然我们也可以用Union Find的方法来做，更简单！

 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]);
    }