LeetCode(32)-Binary Tree Level Order Traversal

题目：

LeetCode Premium SubscriptionProblems Pick OneMock ArticlesDiscussBookfengsehng 102. Binary Tree Level Order Traversal My Submissions QuestionEditorial SolutionTotal Accepted: 98313 Total Submissions: 302608 Difficulty: EasyGiven a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).For example:Given binary tree {3,9,20,#,#,15,7},    3   / \  9  20    /  \   15   7return its level order traversal as:[  [3],  [9,20],  [15,7]]

题意：

• 题意是把一颗二叉树，按照从上到下，，把每一排节点从左到右存进list，最后把所有list存进一个list
• 考虑用递归，设置两个Queue，first和second来存放TreeNode。first来存放当前最下层的一行，second用来存放下一行，first的元素的左右节点赋值给second，把second的元素给first，往下移动
• 考虑first是空的时候，停止，注意判断临时数组tmp是否为空，非空才能存进
-

代码：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> all = new ArrayList<List<Integer>>();        Queue<TreeNode> first = new LinkedList<TreeNode>();        Queue<TreeNode> second = new LinkedList<TreeNode>();        if(root == null){            return all;        }        List<Integer> tmp = new ArrayList<Integer>();        tmp.add(root.val);        all.add(tmp);        first.add(root);        while(!first.isEmpty()){            TreeNode node = first.poll();            if(node.left != null){                second.add(node.left);            }            if(node.right != null){                second.add(node.right);            }            if(first.isEmpty()){                List<Integer> tmp1 = new ArrayList<Integer>();                while(!second.isEmpty()){                    TreeNode n = second.poll();                    first.add(n);                    tmp1.add(n.val);                }                if(tmp1.size() != 0){                    all.add(tmp1);                }                second.clear();            }        }        return all;    }}