the implemention of redblack tree
public class redbalcktree {
private class Node{
private int val;
private int key;
boolean color; //black true
private Node left,right,p;
private int N; //the number of the all nodes of its child nodes and itself
//private int num;//the number
public Node(int val,int key){
this.val = val; this.key = key;
}
}
Node root;
private void leftrotation(Node x){
Node y = x.right;
x.right = y.left;
if(y.left != null){
y.left.p = x;
}
y.p = x.p;
if(x.p == null){
root = y;
}
else if(x == x.p.left){
x.p.left = y;
}
else{x.p.right = y;}
y.left = x;
x.p = y;
}
private void rightrotation(Node y){
Node x = y.left;
y.left = x.right;
if(x.right != null){
x.right.p = y;
}
x.p = y.p;
if(y.p == null){
root = x;
}
else if(y == y.p.left){
y.p.left = x;
}
else{y.p.right = x;}
x.right = y;
y.p = x;
}
public void insert(int val,int key){
Node y = null;
Node x = root;
Node z = new Node(val,key);
if(x == null){root = z;}
while(x != null){
y = x;
if(z.key < x.key){
x = x.left;
}
else{x = x.right;}
}
z.p = y;
if(z.key < y.key){
y.left = z;
}
else{y.right = z;}
z.left = null;
z.right = null;
z.color = false;
inseretfix(z);
}
private void inseretfix(Node z){
Node y;
while (z.p.color == false){
if(z.p == z.p.p.left){
y = z.p.p.right;
if(y.color == false){ //case 1,both z.p and y(uncle) are red
z.p.color = true;
y.color = true;
z.p.p.color = false;
z= z.p.p;
continue;
}
else if(z == z.p.right){ //case 2, z.p is red,uncle is black,z is right,transfer to case 3
z = z.p;
leftrotation(z);
}
z.p.color = true; //case 3, z.p is red,uncle is black,z is left;
z.p.p.color = false;
rightrotation(z.p.p);
}
else{ //set y(uncle) and change rotation
y = z.p.p.left;
if(y.color == false){ //case 1,both z.p and y(uncle) are red
z.p.color = true;
y.color = true;
z.p.p.color = false;
z= z.p.p;
continue;
}
else if(z == z.p.right){ //case 2, z.p is red,uncle is black,z is right,transfer to case 3
z = z.p;
rightrotation(z);
}
z.p.color = true; //case 3, z.p is red,uncle is black,z is left;
z.p.p.color = false;
leftrotation(z.p.p);
}
}
root.color = true;
}
private int get(Node x,int key){
if(x == null){
throw new NullPointerException("have't this key");
}
if(x.key>key){
return get(x.left,key);
}
else if(x.key < key){
return get(x.right,key);
}
else{
return x.val;
}
}
private void transplant(Node u,Node v){ //substitute u as v
if(u.p == null){
root = v;
}
else if(u == u.p.left){
u.p.left = v;
}
else {
u.p.right = v;
}
v.p = u.p;
}
private Node deletemin(Node x){
if(x.left == null){return x.right;}
x.left = deletemin(x.left);
return x;
}
public void delete(int key){
int val = get(root,key);
Node z = new Node(val,key);
Node x;
Node y = z;
boolean original = y.color;
if(z.left == null){
x = z.right;
transplant(z,z.right);
}
else if(z.right == null){
x = z.left;
transplant(z,z.left);
}
else{
y = deletemin(z.right); // next node after z
original = y.color;
x = y.right;
if(y.p == z){
x.p = y;
}
else{
transplant(y,y.right); // put y.right on the original y's position
y.right = z.right;
y.right.p = y;
}
transplant(z,y);
y.left = z.left;
y.left.p = y;
y.color = z.color;
}
if(original == true){
deletefix(x);
}
}
private void deletefix(Node x) {
Node other;
while ((x.color = true) && (x != root)) {
if (x.p.left == x) {
other = x.p.right;
if (other.color == false) {
// Case 1: x's brother is red
other.color = true;
x.p.color = false;
leftrotation(x.p);
other = x.p.right;
}
if ((other.left.color == true) &&
(other.right.color == true) ){
// Case 2: x's black is black and w 's both child are black
other.color = false;
x = x.p;
continue;
} else if (other.right.color == true) {
// Case 3: transfer to case 4
other.left.color = true;
other.color = false;
rightrotation(other);
other = x.p.right;
}
// Case 4: x's black is black,w's right child is red ,left anyway
other.color = x.p.color;
x.p.color = true;
other.right.color = true;
leftrotation(x.p);
x = root;
break;
}
else {
other = x.p.left;
if (other.color == false) {
// Case 1: x's brother is red
other.color = true;
x.p.color = false;
rightrotation(x.p);
other = x.p.right;
}
if ((other.left.color == true) &&
(other.right.color == true) ){
// Case 2: x's black is black and w 's both child are black
other.color = false;
x = x.p;
continue;
} else if (other.right.color == true) {
// Case 3: transfer to case 4
other.left.color = true;
other.color = false;
leftrotation(other);
other = x.p.right;
}
// Case 4: x's black is black,w's right child is red ,left anyway
other.color = x.p.color;
x.p.color = true;
other.right.color = true;
rightrotation(x.p);
x = root;
break;
}
}
}
}
the implemention of redblack tree的更多相关文章
- The easy way to implement a Red-Black tree
Red-Black trees are notorious for being nightmares of pointer manipulation. Instructors will show th ...
- 算法导论学习-RED-BLACK TREE
1. 红黑树(RED-BLACK TREE)引言: ------------------------------------- 红黑树(RBT)可以说是binary-search tree的非严格的平 ...
- [Data Structure] 红黑树( Red-Black Tree ) - 笔记
1. 红黑树属性:根到叶子的路径中,最长路径不大于最短路径的两倍. 2. 红黑树是一个二叉搜索树,并且有 a. 每个节点除了有左.右.父节点的属性外,还有颜色属性,红色或者黑色. b. ( 根属性 ...
- [转]SGI STL 红黑树(Red-Black Tree)源代码分析
STL提供了许多好用的数据结构与算法,使我们不必为做许许多多的重复劳动.STL里实现了一个树结构-Red-Black Tree,它也是STL里唯一实现的一个树状数据结构,并且它是map, multim ...
- TreeMap Red-Black tree
本文以Java TreeMap为例,从源代码层面,结合详细的图解,剥茧抽丝地讲解红黑树(Red-Black tree)的插入,删除以及由此产生的调整过程. 总体介绍 之所以把TreeSet和TreeM ...
- 红黑树(red-black tree)实现记录
https://github.com/xieqing/red-black-tree A Red-black Tree Implementation In C There are several cho ...
- A1135. Is It A Red-Black Tree
There is a kind of balanced binary search tree named red-black tree in the data structure. It has th ...
- PAT 甲级 1135 Is It A Red-Black Tree
https://pintia.cn/problem-sets/994805342720868352/problems/994805346063728640 There is a kind of bal ...
- PAT-A1135. Is It A Red-Black Tree (30)
已知先序序列,判断对应的二叉排序树是否为红黑树.序列中负数表示红色结点,正数表示黑色结点.该序列负数取绝对值后再排序得到的是中序序列.根据红黑树的性质判断它是否符合红黑树的要求.考察了根据先序序列和中 ...
随机推荐
- Python sqlalchemy orm 多外键关联
多外键关联 注:在两个表之间进行多外键链接 如图: 案例: # 创建两张表并添加外键主键 # 调用Column创建字段 加类型 from sqlalchemy import Integer, For ...
- Linux LVM卷组管理
Linux LVM卷组管理 由于传统的磁盘管理不能对磁盘进行磁盘管理,因此诞生了LVM技术,LVM技术最大的特点就是对磁盘进行动态管理. 由于LVM的逻辑卷的大小更改可以进行动态调整,且不会出现丢失数 ...
- eclipse软件仿真操作
1.编写程序代码(以SDRAM为例) 1.1 编写head.s汇编文件 .equ SDRAM_BASE, 0x30000000 .equ MEM_CTL_BASE, 0x48000000 .text ...
- 正确开启Mockjs的三种姿势:入门参考(一)
一.文章初衷 阅读本文章需要注意以下几点: 文章不主要介绍Mockjs的使用语法 文章暂不涉及Mockjs的第三方封装框架 文章会结合以往做过上线项目的方式总结 想主要介绍如何使用Mockjs,是因为 ...
- 安卓外派(Android外派)提供安卓程序员外派业务(北京动点,可签合同)
北京动点飞扬长年提供安卓工程师外派业务. 平均技术情况如下: 1.2~3年以上Android平台开发经验 2.熟练掌握java技术,熟悉面向对象编程设计 3.熟悉Android应用开发框架及Activ ...
- 调研IOS的开发环境的发展演变
一. 关于IOS的开发发展历史: 百度一下,关于这方面的详细资料有很多,在这里就不复制粘贴占用篇幅了. 二. 关于个人搭建IOS开发环境的体验: 本人用的是华硕电脑,window7的操作系统,本来为了 ...
- CSS之user-select——设置标签中的文字是否可被复制
详细介绍请参考 http://www.css88.com/book/css/properties/user-interface/user-select.htm CSS样式 user-select:no ...
- Java实现将文件或者文件夹压缩成zip
最近碰到个需要下载zip压缩包的需求,于是我在网上找了下别人写好的zip工具类.但找了好多篇博客,总是发现有bug.因此就自己来写了个工具类. 这个工具类的功能为: ( ...
- 版本控制——Version Control
版本控制是指对软件开发过程中各种程序代码.配置文件及说明文档等文件变更的管理,是软件配置管理的核心思想之一. 版本控制最主要的功能就是追踪文件的变更.它将什么时候.什么人更改了文件的什么内容等信息忠实 ...
- 机器学习 之XGBoost算法
目录 1.基本知识点简介 2.XGBoost提升树算法 2.1 XGBoost原理 2.2 XGBoost中损失函数的泰勒展开 2.3 XGBoost中正则化项的选定 2.4 最终的目标损失函数及其最 ...