红黑树学习笔记

红黑树所有的红色节点都是左倾斜的，黑色节点为根节点.

代码如下:

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;
		}
	}

}