Java 使用二分搜索树实现映射Map

版权声明：转载请随意！ https://blog.csdn.net/qq_41723615/article/details/89281830

public class BSTMap<K extends Comparable<K>, V> implements Map<K, V> {

    private class Node{
        public K key;
        public V value;
        public Node left, right;

        public Node(K key, V value){
            this.key = key;
            this.value = value;
            left = null;
            right = null;
        }
    }
    //根节点
    private Node root;
    private int size;

    public BSTMap(){
        root = null;
        size = 0;
    }

    @Override
    public int getSize(){return size; }

    @Override
    public boolean isEmpty(){ return size == 0;}

    // 向二分搜索树中添加新的元素(key, value)
    @Override
    public void add(K key, V value){
        //递归调用
        root = add(root, key, value);
    }

    // 向以node为根的二分搜索树中插入元素(key, value)，递归算法
    // 返回插入新节点后二分搜索树的根
    private Node add(Node node, K key, V value){

        if(node == null){
            size ++;
            return new Node(key, value);
        }
        //由于二分搜索树是有序的，所以需要比较判断
        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
            //替换
            node.value = value;
        }
        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);
        }
    }

    @Override
    public boolean contains(K key){
        return getNode(root, key) != null;
    }

    @Override
    public V get(K key){

        Node node = getNode(root, key);
        return node == null ? null : node.value;
    }

    @Override
    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
    @Override
    public V remove(K key){

        Node node = getNode(root, key);
        //如果节点不为null
        if(node != null){
            //递归调用
            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;
        }
    }

测试方法：

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());

            BSTMap<String, Integer> map = new BSTMap<>();
            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();
    }

结果：