1每个节点是红色或者黑色
2根节点是黑色
3每一个叶子节点（最后的空节点）是黑色
4如果一个节点是红色，那么他的孩子节点都是黑色，红色节点都是向左倾斜
5从任意一个节点到叶子节点，经过的黑色节点是一样的
也满足二分搜索树的性质，是绝对平衡的树，从根节点到任意节点都是经过相同的层数。
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颜色翻转
颜色翻转
import java.util.ArrayList;
public class RBTree<K extends Comparable<K>, V> {
private static final boolean RED = true;
private static final boolean BLACK = false;private class Node{
public K key;
public V value;
public Node left, right;
public boolean color;public Node(K key, V value){
this.key = key;
this.value = value;
left = null;
right = null;
color = RED;
}
}private Node root;
private int size;public RBTree(){
root = null;
size = 0;
}public int getSize(){
return size;
}public boolean isEmpty(){
return size == 0;
}// 判断节点node的颜色
private boolean isRed(Node node){
if(node == null)
return BLACK;
return node.color;
}// node x
// / \ 左旋转 / \
// T1 x ---------> node T3
// / \ / \
// T2 T3 T1 T2
private Node leftRotate(Node node){Node x = node.right;
// 左旋转
node.right = x.left;
x.left = node;x.color = node.color;
node.color = RED;return x;
}// node x
// / \ 右旋转 / \
// x T2 -------> y node
// / \ / \
// y T1 T1 T2
private Node rightRotate(Node node){Node x = node.left;
// 右旋转
node.left = x.right;
x.right = node;x.color = node.color;
node.color = RED;return x;
}// 颜色翻转
private void flipColors(Node node){node.color = RED;
node.left.color = BLACK;
node.right.color = BLACK;
}// 向红黑树中添加新的元素(key, value)
public void add(K key, V value){
root = add(root, key, value);
root.color = BLACK; // 最终根节点为黑色节点
}// 向以node为根的红黑树中插入元素(key, value)，递归算法
// 返回插入新节点后红黑树的根
private Node add(Node node, K key, V value){if(node == null){
size ++;
return new Node(key, value); // 默认插入红色节点
}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;if (isRed(node.right) && !isRed(node.left))
node = leftRotate(node);if (isRed(node.left) && isRed(node.left.left))
node = rightRotate(node);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;
}
}public static void main(String[] args){
System.out.println("Pride and Prejudice");
ArrayList<String> words = new ArrayList<>();
if(FileOperation.readFile("pride-and-prejudice.txt", words)) {
System.out.println("Total words: " + words.size());RBTree<String, Integer> map = new RBTree<>();
for (String word : words) {
if (map.contains(word))
map.set(word, map.get(word) + 1);
else
map.add(word, 1);
}System.out.println("Total different words: " + map.getSize());
System.out.println("Frequency of PRIDE: " + map.get("pride"));
System.out.println("Frequency of PREJUDICE: " + map.get("prejudice"));
}System.out.println();
}
}