package wck.sort;/**
* Created on 18/9/13.
*/
/**
* @program: demo
* @description: 红黑树实现
* @author: wck
* @create: 2018-09-13 11:07
**/
public class RBT <K extends Comparable<K>, V> {
private final static boolean RED = true;
private final static boolean BLACK = false;
private class Node{
public K key;
public V value;
public Node left, right;
private boolean color; //定义红黑树颜色
public Node(K key, V value){
this.key = key;
this.value = value;
left = null;
right = null;
//向上融合的过程，临时的四节点向上的父亲节点融合，而且因为在左边，肯定比父节点小。
//所以向上融合的元素都是红色的节点。－－－极端情况下，融合到最上头，根节点为红，所以要变色；
color =RED;
}
}
private Node root;
private int size;
public RBT(){
root = null;
size = 0;
}
public int getSize(){
return size;
}
public boolean isEmpty(){
return size == 0;
}
//判断节点颜色
private boolean isRed(Node node){
if(node == null)
return BLACK;
return node.color;
}
//二节点添加元素左旋过程：
// node right
// / \ 左旋转 / \
// T1 right ---------> node T3
// / \ / \
// T2 T3 T1 T2
private Node leftRotate(Node node){
Node right = node.right;
node.right = right.left;
right.left = node;
right.color = node.color;
node.color = RED;
return right;
}
//三节点加颜色翻转
private void flipColors(Node node){
node.right.color =BLACK;
node.left.color = BLACK;
node.color = RED;
}
//三节点添加元素右旋过程：
// node x
// / \ 右旋转 / \
// x T2 -------> y node
// / \ / \
// y T1 T1 T2
private Node rightRotate(Node node){
Node left = node.left;
node.left =left.right;
left.right = node;
left.color = BLACK;
left.color = RED;
return left;
}
// 向二分搜索树中添加新的元素(key, value)
public void add(K key, V value){
root = add(root, key, value);
root.color = BLACK;
}
// 向以node为根的二分搜索树中插入元素(key, value)，递归算法
// 返回插入新节点后二分搜索树的根
private Node add(Node node, K key, V value){
if(node == null){
size ++;
return new Node(key, value);//默认为红赛～
}
if(key.compareTo(node.key) < 0)
node.left = add(node.left, key, value);
else if(key.compareTo(node.key) > 0)
node.right = add(node.right, key, value);
else // key.compareTo(node.key) == 0
node.value = value;
//左旋
if(isRed(node.right) && !isRed(node.left))
node = leftRotate(node);
//右旋
if(isRed(node.left) && isRed(node.left.left))
node = rightRotate(node);
//翻转
if(isRed(node.left) && isRed(node.right))
flipColors(node);
return node;
}
// 返回以node为根节点的二分搜索树中，key所在的节点
private Node getNode(Node node, K key){
if(node == null)
return null;
if(key.equals(node.key))
return node;
else if(key.compareTo(node.key) < 0)
return getNode(node.left, key);
else // if(key.compareTo(node.key) > 0)
return getNode(node.right, key);
}
public boolean contains(K key){
return getNode(root, key) != null;
}
public V get(K key){
Node node = getNode(root, key);
return node == null ? null : node.value;
}
public void set(K key, V newValue){
Node node = getNode(root, key);
if(node == null)
throw new IllegalArgumentException(key + " doesn't exist!");
node.value = newValue;
}
// 返回以node为根的二分搜索树的最小值所在的节点
private Node minimum(Node node){
if(node.left == null)
return node;
return minimum(node.left);
}
// 删除掉以node为根的二分搜索树中的最小节点
// 返回删除节点后新的二分搜索树的根
private Node removeMin(Node node){
if(node.left == null){
Node rightNode = node.right;
node.right = null;
size --;
return rightNode;
}
node.left = removeMin(node.left);
return node;
}
// 从二分搜索树中删除键为key的节点
public V remove(K key){
Node node = getNode(root, key);
if(node != null){
root = remove(root, key);
return node.value;
}
return null;
}
private Node remove(Node node, K key){
if( node == null )
return null;
if( key.compareTo(node.key) < 0 ){
node.left = remove(node.left , key);
return node;
}
else if(key.compareTo(node.key) > 0 ){
node.right = remove(node.right, key);
return node;
}
else{ // key.compareTo(node.key) == 0
// 待删除节点左子树为空的情况
if(node.left == null){
Node rightNode = node.right;
node.right = null;
size --;
return rightNode;
}
// 待删除节点右子树为空的情况
if(node.right == null){
Node leftNode = node.left;
node.left = null;
size --;
return leftNode;
}
// 待删除节点左右子树均不为空的情况
// 找到比待删除节点大的最小节点, 即待删除节点右子树的最小节点
// 用这个节点顶替待删除节点的位置
Node successor = minimum(node.right);
successor.right = removeMin(node.right);
successor.left = node.left;
node.left = node.right = null;
return successor;
}
}
}
算法基础之--红黑树实现
猜你喜欢
转载自blog.csdn.net/qq_36866808/article/details/82687936
今日推荐
周排行