1.. 图解2-3树维持绝对平衡的原理：
2.. 红黑树与2-3树是等价的
3.. 红黑树的特点
- 简要概括如下：
- 所有节点非黑即红；根节点为黑；NULL节点为黑；红节点孩子为黑；黑平衡
4.. 实现红黑树的业务逻辑
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();
}
}
5.. 向红黑树中添加新元素之后需要进行维护的过程示意图