算法基礎之--紅黑樹實現

package wck.sort;/**
 * Created on 18/9/13.
 */

/**
 * @program: demo
 * @description: 紅黑樹實現
 * @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; //定義紅黑樹顏色

            public Node(K key, V value){
                this.key = key;
                this.value = value;
                left = null;
                right = null;
                //向上融合的過程，臨時的四節點向上的父親節點融合，而且因爲在左邊，肯定比父節點小。
                //所以向上融合的元素都是紅色的節點。－－－極端情況下，融合到最上頭，根節點爲紅，所以要變色；
                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;
        }

        //判斷節點顏色
        private boolean isRed(Node node){
            if(node == null)
                return BLACK;
            return node.color;
        }

        //二節點添加元素左旋過程：
        //   node                     right
        //  /   \     左旋轉         /  \
        // 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;
        }

        //三節點加顏色翻轉
        private void flipColors(Node node){
            node.right.color =BLACK;
            node.left.color = BLACK;
            node.color = RED;
        }

        //三節點添加元素右旋過程：
        //     node                   x
        //    /   \     右旋轉       /  \
        //   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;
        }


        // 向二分搜索樹中添加新的元素(key, value)
        public void add(K key, V value){
            root = add(root, key, value);
            root.color = BLACK;
        }

        // 向以node爲根的二分搜索樹中插入元素(key, value)，遞歸算法
        // 返回插入新節點後二分搜索樹的根
        private Node add(Node node, K key, V value){

            if(node == null){
                size ++;
                return new Node(key, value);//默認爲紅賽～
            }

            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);
            //翻轉
            if(isRed(node.left) && isRed(node.right))
                flipColors(node);

            return node;
        }

        // 返回以node爲根節點的二分搜索樹中，key所在的節點
        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爲根的二分搜索樹的最小值所在的節點
        private Node minimum(Node node){
            if(node.left == null)
                return node;
            return minimum(node.left);
        }

        // 刪除掉以node爲根的二分搜索樹中的最小節點
        // 返回刪除節點後新的二分搜索樹的根
        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;
        }

        // 從二分搜索樹中刪除鍵爲key的節點
        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

                // 待刪除節點左子樹爲空的情況
                if(node.left == null){
                    Node rightNode = node.right;
                    node.right = null;
                    size --;
                    return rightNode;
                }

                // 待刪除節點右子樹爲空的情況
                if(node.right == null){
                    Node leftNode = node.left;
                    node.left = null;
                    size --;
                    return leftNode;
                }

                // 待刪除節點左右子樹均不爲空的情況

                // 找到比待刪除節點大的最小節點, 即待刪除節點右子樹的最小節點
                // 用這個節點頂替待刪除節點的位置
                Node successor = minimum(node.right);
                successor.right = removeMin(node.right);
                successor.left = node.left;

                node.left = node.right = null;

                return successor;
            }
        }

}