很巧妙的，又学习了一种层次遍历的方法，就是说根据当前的队列的长度去遍历，遍历的当前队列的长度就是该层次的节点个数。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> levelOrderList = new ArrayList<>();
        if(root == null) return levelOrderList;
        Queue<TreeNode> queue = new LinkedList<>();    //用于做队列，遍历整棵树
        List<Integer> list = new ArrayList<>();     // 用于存储用队列遍历树的val
        queue.add(root);
        while(queue.size()!=0) {
        //    System.out.println("当前队列的长度" + queue.size());
            int l = queue.size();
            for(int i=0;i<l; i++) {
                TreeNode front = queue.poll();   //LinkedList的poll方法可以取出第一个元素的值。
                list.add(front.val);
                if(front.left!=null) queue.add(front.left);
                if(front.right!=null) queue.add(front.right);
            }
/*             System.out.print("当前放入的元素为");
            Iterator iterator = list.iterator();
            while(iterator.hasNext()){
                System.out.print(iterator.next() + " ");
            }
            System.out.println(); */
            levelOrderList.add(list);
            list = new ArrayList<Integer>();
        }
        return levelOrderList;
    }
}