LeetCode_107. Binary Tree Level Order Traversal II
107. Binary Tree Level Order Traversal II
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7]
,
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
package leetcode.easy; /**
* Definition for a binary tree node. public class TreeNode { int val; TreeNode
* left; TreeNode right; TreeNode(int x) { val = x; } }
*/
public class BinaryTreeLevelOrderTraversalII {
// DFS solution:
public java.util.List<java.util.List<Integer>> levelOrderBottom1(TreeNode root) {
java.util.List<java.util.List<Integer>> wrapList = new java.util.LinkedList<java.util.List<Integer>>();
levelMaker(wrapList, root, 0);
return wrapList;
} private static void levelMaker(java.util.List<java.util.List<Integer>> list, TreeNode root, int level) {
if (root == null) {
return;
}
if (level >= list.size()) {
list.add(0, new java.util.LinkedList<Integer>());
}
levelMaker(list, root.left, level + 1);
levelMaker(list, root.right, level + 1);
list.get(list.size() - level - 1).add(root.val);
} // BFS solution:
public java.util.List<java.util.List<Integer>> levelOrderBottom2(TreeNode root) {
java.util.Queue<TreeNode> queue = new java.util.LinkedList<TreeNode>();
java.util.List<java.util.List<Integer>> wrapList = new java.util.LinkedList<java.util.List<Integer>>();
if (root == null) {
return wrapList;
}
queue.offer(root);
while (!queue.isEmpty()) {
int levelNum = queue.size();
java.util.List<Integer> subList = new java.util.LinkedList<Integer>();
for (int i = 0; i < levelNum; i++) {
if (queue.peek().left != null) {
queue.offer(queue.peek().left);
}
if (queue.peek().right != null) {
queue.offer(queue.peek().right);
}
subList.add(queue.poll().val);
}
wrapList.add(0, subList);
}
return wrapList;
} @org.junit.Test
public void test() {
TreeNode tn11 = new TreeNode(3);
TreeNode tn21 = new TreeNode(9);
TreeNode tn22 = new TreeNode(20);
TreeNode tn33 = new TreeNode(15);
TreeNode tn34 = new TreeNode(7);
tn11.left = tn21;
tn11.right = tn22;
tn21.left = null;
tn21.right = null;
tn22.left = tn33;
tn22.right = tn34;
tn33.left = null;
tn33.right = null;
tn34.left = null;
tn34.right = null;
System.out.println(levelOrderBottom1(tn11));
System.out.println(levelOrderBottom2(tn11));
}
}
LeetCode_107. Binary Tree Level Order Traversal II的更多相关文章
- 35. Binary Tree Level Order Traversal && Binary Tree Level Order Traversal II
Binary Tree Level Order Traversal OJ: https://oj.leetcode.com/problems/binary-tree-level-order-trave ...
- Binary Tree Level Order Traversal,Binary Tree Level Order Traversal II
Binary Tree Level Order Traversal Total Accepted: 79463 Total Submissions: 259292 Difficulty: Easy G ...
- LeetCode之“树”:Binary Tree Level Order Traversal && Binary Tree Level Order Traversal II
Binary Tree Level Order Traversal 题目链接 题目要求: Given a binary tree, return the level order traversal o ...
- 102/107. Binary Tree Level Order Traversal/II
原文题目: 102. Binary Tree Level Order Traversal 107. Binary Tree Level Order Traversal II 读题: 102. 层序遍历 ...
- 【LeetCode】107. Binary Tree Level Order Traversal II (2 solutions)
Binary Tree Level Order Traversal II Given a binary tree, return the bottom-up level order traversal ...
- 63. Binary Tree Level Order Traversal II
Binary Tree Level Order Traversal II My Submissions QuestionEditorial Solution Total Accepted: 79742 ...
- [LeetCode] Binary Tree Level Order Traversal II 二叉树层序遍历之二
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
- 【Binary Tree Level Order Traversal II 】cpp
题目: Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from ...
- Binary Tree Level Order Traversal II 解题思路
思路: 与Binary Tree Level Order Traversal I 几乎一样.只是最后将结果存放在栈里,然后在栈里再传给向量即可. 再次总结思路: 两个queue,先把第一个放进q1,循 ...
随机推荐
- 运维CMDB建设思路
在我们日常的运维工作中,面对着大量的基础设施和软件服务,该如何管理?这个管理的原则又是什么?粒度该如何控制?我们是否可以建立一个统一的标准模型来管理以上对象?管理过程中,如何降低人力成本?资源对象的生 ...
- springcloud注册中心Eureka<英 [juəˈri:kə]>的基本搭建
1.http://start.spring.io搭建基本的springboot环境,版本用的是1.5.10 2.在pom文件中添加注册中心的jar包和springcloud的jar包 <!-- ...
- 线程(C++11)
不同平台.框架.语言所使用的线程函数不同.对于初学者选择一种适合自己的,用熟用透即可. Windows中,CreateThread() Linux中,pthread_create() MFC框架中,A ...
- Linux 硬件软件时间同步
同步BIOS时钟,强制把系统时间写入CMOS clock --show 查看硬件时间clock -w 强制把系统时间写入CMOSclock --show 查看硬件时间reboot ...
- learning scala type alise
How to use type alias to name a Tuple2 pair into a domain type called CartItem type CartItem[Donut, ...
- leetcode解题报告(16):Move Zeroes
描述 Given an array nums, write a function to move all 0's to the end of it while maintaining the rela ...
- 2-set奶牛议会
Description 由于对Farmer John的领导感到极其不悦,奶牛们退出了农场,组建了奶牛议会.议会以“每头牛 都可以获得自己想要的”为原则,建立了下面的投票系统: M只到场的奶牛 (1 & ...
- 使用Keras训练神经网络备忘录
小书匠深度学习 文章太长,放个目录: 1.优化函数的选择 2.损失函数的选择 2.2常用的损失函数 2.2自定义函数 2.1实践 2.2将损失函数自定义为网络层 3.模型的保存 3.1同时保持结构和权 ...
- 前端逼死强迫症之DOM
Dom:document.相当于把所有的html文件,转换成了文档对象. 之前说过:html-裸体的人:css-穿上衣服:js-让人动起来. 让人动起来,就得先找到他,再修改它内容或属性. 找到标签 ...
- python ros 四元数转欧拉角
#! /usr/bin/python import PyKDL import rospy from sensor_msgs.msg import Imu from nav_msgs.msg impor ...