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

  1. The easy way to implement a Red-Black tree

    Red-Black trees are notorious for being nightmares of pointer manipulation. Instructors will show th ...

  2. 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 ...

  3. [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 ...

  4. [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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. PAT 甲级 1135 Is It A Red-Black Tree

    https://pintia.cn/problem-sets/994805342720868352/problems/994805346063728640 There is a kind of bal ...

  9. 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

  1. python 竖排文本

    新建目录train,并将目录data和data1复制到train下 python test data/,data1/ 目录data和data1中包含很多文件,文件中内容都是以空格分隔,将所有文件内容都 ...

  2. 在网页中运用统计Web Service接口

    (2017-02-10 银河统计) 在"统计随机数及临界值Web Service接口"一文中介绍了常用统计分布四类Web Service接口(随机数.分位数.密度函数和累积分布函数 ...

  3. 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 ...

  4. JS(JavaScript)的初了解3(更新中···)

    1. {} 在JS中我们把它叫代码块.如果代码块里的内容没有执行完,语句不会向下执行. 代码块是一个独立的整体.如果JS中某一条语句出错,那么就会在此终止不会向下执行. 2. 循环语句 循环,就是对一 ...

  5. ArchLinux 无密码Samba 配置

    安装上迅雷后就必须把硬盘挂上去开个Samba,然后边下电影边看,这就是一个Nas了. 开搞! pacman -S samba 安装samba,然后进入/etc/samba/目录 vim smb.con ...

  6. tp剩余未验证内容-7

    bash脚本中 的 set -e表示 exit immediately if a simple command returns a non-zero value.主要是为了防止错误被忽略.会被立即退出 ...

  7. sqlserver 判断字符串是否是数字

    https://www.cnblogs.com/zjfblog/p/5625202.html sql2005有个函数ISNUMERIC(expression)函数:当expression为数字时,返回 ...

  8. Ubuntu 14.04 更新 setuptools 至 19.2 版本

    参考: Error: "No module named _markerlib" when installing some packages on virtualenv Ubuntu ...

  9. 关于var time = +new Date;

    文章地址:https://www.cnblogs.com/Raoh/p/4212075.html

  10. noip模拟ernd

    [题目背景]解决了第一题之后,你打开了第二题.这是一道关于树的题目,你很快想出了一个复杂度和树的直径有关的算法,可以通过所有的数据.不过,你的做法常数似乎有点大.为了更好地卡常,你决定构造一些数据来检 ...