第三十三篇 玩转数据结构——红黑树(Read Black Tree)

 

 

 

1.. 图解2-3树维持绝对平衡的原理:

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2.. 红黑树与2-3树是等价的

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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

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posted @ 2018-07-14 13:34  XueZou  阅读(484)  评论(0编辑  收藏  举报