数据结构平衡二叉树

Posted 2022-03-14 写Bug的渣渣高

/**
 * 平衡二叉树
 *  对于 BST 节点元素，一定要实现 Comparable 接口
 * @param <E>
 */
public class BST<E extends Comparable<E>> 

    /**
     * 二叉树的节点
     */
    private class Node
        public E e;
        public Node left ,right;
        public Node(E e)
            this.e = e;
            left = null;
            right = null;
        
    

    private Node root;
    private int size;

    public BST()
        root = null;
        size =0;
    

    public void add(E e)
//        if (root == null)
//            root = new Node(e);
//            size++;
//        else 
        root = add(root,e);
//        
    

    /**
     * 需要以node为跟然后使用递归插入元素
     * @param node
     * @param e
     */
    private Node add(Node node ,E e )
//      不包括重复情况
//        if (e.equals(node.e))
//            return;
//        else if (e.compareTo(node.e) < 0 && node.left == null)
//            node.left = new Node(e);
//            size++;
//            return;
//        else if (e.compareTo(node.e) > 0 && node.right ==null)
//            node.right = new Node(e);
//            size++;
//            return;
//        
//        if (e.compareTo(node.e) < 0)
//            add(node.left,e);
//        else 
//            add(node.right,e);
//        
        /**
         * 使用的时递归的方法来添加如果，如果这个节点为空，那么根据值创建一个node并返回
         */
        if (node == null)
            size++;
            return new Node(e);
        if (e.compareTo(node.e) < 0)
            node.left = add(node.left,e);
        else if (e.compareTo(node.e) > 0)
            node.right = add(node.right,e);
        
        return node;
    

    public boolean contains(E e)
        return contains(root,e);
    

//    看以node为跟的二分搜搜树是否包含元素e
    private boolean contains(Node node ,E e)
        if (node == null)
            return false;
        
        if (e.compareTo(node.e) == 0)
            return true;
        else if (e.compareTo(node.e) < 0)
            contains(node.left,e);
        else 
            contains(node.right,e);
        
        return false;
    

    public void preOrder()
        preOrder(root);
    

    private void preOrder(Node node)
        if (node == null)
            return;
        
        System.out.println(node.e);
        preOrder(node.left);
        preOrder(node.right);
    

    /**
     * 非递归的前序遍历
     */
    public void preOrderNR()
        Stack<Node> stack = new Stack<Node>();
        stack.push(root);
        while (!stack.isEmpty()) 
            Node cur = stack.pop();
            System.out.println(cur.e);
            if (cur.left != null)
                stack.push(cur.left);
            if (cur.right != null)
                stack.push(cur.right);
        
    

    /**
     * 先释放所有孩子节点，再释放节点本身
     */
    public void postOrder()
        postOrder(root);
    

    private void postOrder(Node node)
        if (node == null)
            return;
        
        postOrder(node.left);
        postOrder(node.right);
        System.out.println(node.e);
    

    /**
     * 中序遍历就是顺序遍历
     */
    public void inOrder()
        inOrder(root);
    

    private void inOrder(Node node)
        if (node == null)
            return;
        
        inOrder(node.left);
        System.out.println(node.e);
        inOrder(node.right);
    

    public E minMum()
        if (size == 0)
            throw new IllegalArgumentException("二分搜索树为空");
        
        return minMum(root).e;
    

    private Node minMum(Node node)
        if (node.left == null)
            return node;
        
        return minMum(node.left);
    

    /**
     * 如何删除？？
     *      如果最小值是叶子，直接删
     *      否则就删了该节点，然后把该节点右子树接到该节点双亲节点
     * @return
     */
    public E removeMin()
        E ret = minMum();
        root = removeMin(root);
        return ret;
    

    private Node removeMin(Node node)
        if (node.left == null)
//          这里右子树可能为空
            Node rightNode = node.right;
            node.right = null;
            size --;
            return rightNode;
        
        node.left = removeMin(node.left);
        return node;
    

    public E removeMax()
        E ret = maxMum();
        root = removeMin(root);
        return ret;
    

    private Node removeMax(Node node)
        if (node.right == null)
            Node rightNode = node.left;
            node.right = null;
            size --;
            return rightNode;
        
        node.right = removeMin(node.right);
        return node;
    

    public E maxMum()
        if (size == 0)
            throw new IllegalArgumentException("二分搜索树为空");
        
        return maxMum(root).e;
    

    private Node maxMum(Node node)
        if (node.right == null)
            return node;
        
        return maxMum(node.right);
    

    public void remove(E e)
        root = remove(root,e);
    

    private Node remove(Node node,E e)
        if (node == null)
            return null;
        
        if (e.compareTo(node.e) < 0)
            node.left = remove(node.left,e);
        else if (e.compareTo(node.e) > 0)
            node.right = remove(node.right,e);
        else  //e == node.e
            if (node.left == null) 
                Node rightNode = node.right;
                node.right = null;
                size--;
                return rightNode;
            
            if (node.right == null)
                Node leftNode = node.left;
                node.left = null;
                size --;
                return leftNode;
            
        
//      待删除节点均不为空,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
        Node successor = minMum(node.right);
        successor.right = removeMin(node.right);
        successor.left = node.left;
        node.left = node.right = null;
        return successor;
    

    /**
     * 广度优先遍历.层序遍历，需要使用到队列
     */
    public void levelOrder()
        Queue<Node> q = new LinkedList<>();
        q.add(root);
        while (!q.isEmpty())
            Node cur = q.remove();
            System.out.println(cur.e);
            if (cur.left != null)
                q.add(cur.left);
            
            if (cur.right != null)
                q.add(cur.right);
            
        

    

    public int size()
        return size;
    

    public boolean isEmpty()
        return size==0;
    



    @Override
    public String toString() 
        StringBuilder res = new StringBuilder();
        generateBSTString(root,0,res);
        return res.toString();
    

    private void generateBSTString(Node root, int i, StringBuilder res) 
        if (root == null)
            res.append(generateDepthString(i)+"null\\n");
            return;
        
        res.append(generateDepthString(i)+root.e+"\\n");
        generateBSTString(root.left,i+1,res);
        generateBSTString(root.right,i+1,res);
    

    private String generateDepthString(int i) 
        StringBuilder res = new StringBuilder();
        for (int j = 0; j < i; j++) 
            res.append("--");
        
        return res.toString();