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
- mailto标签来调用邮箱客户端
最近项目需要使用mailto标签来调用客户端,并且把邮件模板填到客户端. mailto 的用法: a标签直接调用: <a href="mailto:example@qq.com?cc= ...
- C++——构造和析构函数
现在学习进入第三阶段,对c++要有更深入的学习,关于构造函数和析构函数这一块需要总结一下,来深刻理解这两个函数的意义. 什么是构造函数和析构函数呢呢?听着就很高大上,但是要从心里藐视它.就像自然万物有 ...
- 2019年java技术大盘点
福州SEO:2019年互联网企业在Java开发中有哪些主流.热门的IT技术呢,下面让我们来看一下. 微服务技术 微服务架构主要有:Spring Cloud. Dubbo. Dubbox等,以 Dubb ...
- mongoose 5.0 链接数据库 代码保存
const mongoose = require('mongoose'); const dbSrc = 'mongodb://localhost/douban-trailer' mongoose.Pr ...
- theme-sodareload sublime编辑器主题插件还不错,不是语法高亮
theme-sodareload sublime编辑器主题还不错,不是语法高亮
- Codeforces Round #555 (Div. 3) C1,C2【补题】
D1:思路:L,R指针移动,每次选最小的即可. #include<bits/stdc++.h> using namespace std; #define int long long #de ...
- codevs 4244 平衡树练习
二次联通门 : codevs 4244 平衡树练习 Splay实测指针占用空间大约是数组的3倍, 且时间上也慢了差不多1s 数组版评测记录如下 指针版评测记录如下 以上数据仅限这一个题, 对于 ...
- ROS与树莓派的结合
从零开始学树莓派和ROS 今天写下自己的第一篇博客,记录一下自己的学习历程和学习过程中碰到的各种小问题,供同道者参阅和自己以后回顾用 ,水平不高,我就放开手写吧,反正也不会有人看. 我现在在做毕业设计 ...
- Allure自动化测试报告之修改allure测试报告名称
1.从github获取allure代码 https://github.com/allure-framework/allure2 2.安装gradle,用于打包jar brew install grad ...
- EXTJS框架-入门实例
extjs框架是一个JavaScript框架,可以渲染出丰富的控件 实例: 代码: <html> <head> <title>test</title> ...