力扣 二叉树的层序遍历

题目：https://leetcode-cn.com/problems/binary-tree-level-order-traversal/

需要注意的点：

• Java的Queue相关的内容，例如，实例化时其引用一般是指向LinkedList对象，还有offer(),peek(),poll()，isEmpty()方法。

代码：一个是加了个flag节点标记；一个是记录size

/**
 * 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>> res = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>(); 
        TreeNode tmp = new TreeNode();
        TreeNode flag = new TreeNode();
        if(root == null){
            return res;
        }
        q.offer(root);
        q.offer(flag);
        while(true){
            List<Integer> l = new ArrayList<>();
            while(!q.isEmpty()){
                tmp = q.peek();
                l.add(tmp.val);
                if(tmp.left != null){
                    q.offer(tmp.left);
                }
                if(tmp.right != null){
                    q.offer(tmp.right);
                }
                q.poll();
                if(q.peek() == flag){
                    q.poll();
                    if(q.isEmpty()){
                        break;
                    }
                    q.offer(flag);
                    break;
                } 
            }
            res.add(l);
            if(q.isEmpty()){
                break;
            }
        }
        return res;
    }
}

/**
 * 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>> res = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>(); 
        TreeNode tmp = new TreeNode();
        if(root == null){
            return res;
        }
        q.offer(root);
        while(q.isEmpty() == false){
            int cur_size = q.size();
            List<Integer> l = new ArrayList<>();
            for(int i = 0; i < cur_size; i ++){
                tmp = q.poll();
                l.add(tmp.val);
                if(tmp.left != null){
                        q.offer(tmp.left);
                    }
                if(tmp.right != null){
                    q.offer(tmp.right);
                }
            }
            res.add(l);
        }
        
        return res;
    }
}