数据结构之二分搜索树（1）——手写二分搜索树

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

public class BST> {

    private class Node {
        public E e;

        private 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 int size() {
        return size;
    }

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

    /**
     * 向二分搜索树中添加一个新的元素e
     *
     * @param e
     */
    public void add(E e) {
        root = add(root, e);
    }

    //向以node为根的二分搜索树中插入元素E,递归算法
    //返回插入新节点后二分搜索树的根
    private Node add(Node node, E e) {
        if (node == null) {
            size++;
            return new Node(e);
        }
        if (e.compareTo(node.e) < 0) {
            //插入左子树
            node.left = add(node.left, e);
        } else if (e.compareTo(node.e) > 0) {
            //插入右子树
            node.right = add(node.right, e);
        }
        return node;
    }

    /**
     * 看二分搜素树中是否包含元素e
     *
     * @param e
     * @return
     */
    public boolean contains(E e) {
        return contains(root, e);
    }

    /**
     * 看以node为根的二分搜索树中是否包含元素e,递归算法
     *
     * @param node
     * @param e
     * @return
     */
    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) {
            return contains(node.left, e);
        } else {
            return contains(node.right, e);
        }
    }

    //二分搜索树的前序遍历
    public void preOrder() {
        preOrder(root);
    }

    //二分搜索树的前序遍历(非递归方式)
    public void preOrderNR() {
        Stack stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            Node cur = stack.pop();
            System.out.println(cur.e);
            if (cur.right != null) {
                stack.push(cur.right);
            }
            if (cur.left != null) {
                stack.push(cur.left);
            }

        }
    }


    //二分搜索树的中序遍历
    public void inOrder() {
        inOrder(root);
    }

    //二分搜索树的后序遍历
    public void postOrder() {
        postOrder(root);
    }

    public void levelOrder() {
        Queue 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);
            }

        }

    }

    /**
     * 前序遍历以Node为根的二分搜索树，递归算法
     *
     * @param node
     */
    private void preOrder(Node node) {
        if (node == null) {
            return;
        }
        System.out.println(node.e);
        preOrder(node.left);
        preOrder(node.right);
    }

    /**
     * 中序遍历以Node为根的二分搜索树，递归算法
     *
     * @param node
     */
    private void inOrder(Node node) {
        if (node == null) {
            return;
        }
        inOrder(node.left);
        System.out.println(node.e);
        inOrder(node.right);
    }

    /**
     * 后序遍历以Node为根的二分搜索树，递归算法
     *
     * @param node
     */
    private void postOrder(Node node) {
        if (node == null) {
            return;
        }
        postOrder(node.left);
        postOrder(node.right);
        System.out.println(node.e);
    }

    /**
     * 寻找二分搜索树中的最小元素
     *
     * @return
     */
    public E minmun() {
        if (size == 0) {
            throw new IllegalArgumentException("BST IS EMPTY!");
        }
        Node node = minmun(root);
        return node.e;
    }

    private Node minmun(Node node) {
        if (node != null && node.left == null) {
            return node;
        }
        return minmun(node.left);
    }

    /**
     * 寻找二分搜索树中的最大元素
     *
     * @return
     */
    public E maxmun() {
        if (size == 0) {
            throw new IllegalArgumentException("BST IS EMPTY!");
        }
        Node node = maxmun(root);
        return node.e;
    }

    private Node maxmun(Node node) {
        if (node != null && node.right == null) {
            return node;
        }
        return maxmun(node.right);
    }

    /**
     * 从二分搜索树中删除最小值所在的节点，返回最小值
     *
     * @return
     */
    public E removeMin() {
        E ret = minmun();
        root = removeMin(root);
        return ret;
    }

    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最大值所在的节点，返回最大值
     *
     * @return
     */
    public E removeMax() {
        E ret = maxmun();
        root = removeMax(root);
        return ret;
    }

    private Node removeMax(Node node) {
        if (node.right == null) {
            Node leftNode = node.left;
            node.left = null;
            size--;
            return leftNode;
        }
        node.right = removeMax(node.right);
        return node;
    }

    /**
     * 从二分搜索树中删除元素为e的节点
     *
     * @param e
     * @return
     */
    public void remove(E e) {
        root = remove(root, e);
    }

    /**
     * 从二分搜索树中删除元素为e的节点
     *
     * @param node
     * @param e
     * @return
     */
    private Node remove(Node node, E e) {
        if (node == null) {
            return null;
        }
        if (e.compareTo(node.e) < 0) {
            node.left = remove(node.left, e);
            return node;
        } else if (e.compareTo(node.e) > 0) {
            node.right = remove(node.right, e);
            return node;
        } else {
            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 = minmun(node.right);
        successor.right = removeMin(node.right);
        successor.left = node.left;
        node.left = node.left = null;
        return successor;

    }

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

    /**
     * 生成以Node为节点，深度为depth的描述二叉树的字符串
     *
     * @param node
     * @param depth
     * @param res
     */
    private void generateBSTStr(Node node, int depth, StringBuilder res) {
        if (node == null) {
            res.append(generateDepthString(depth) + "null\n");
            return;
        }
        res.append(generateDepthString(depth) + node.e + "\n");
        generateBSTStr(node.left, depth + 1, res);
        generateBSTStr(node.right, depth + 1, res);
    }

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

 

import java.util.ArrayList;
import java.util.Random;

public class Main {

    public static void main(String[] args) {
        BST bst = new BST<>();
        /*int[] nums = {5, 3, 6, 8, 4, 2};
        for (int num : nums) {
            bst.add(num);
        }
       // bst.preOrder();

        System.out.println();
       // bst.inOrder();
        //bst.postOrder();
        //System.out.println(bst);
        //bst.preOrderNR();
        bst.levelOrder();*/
        Random random = new Random();
        int n=1000;
        for (int i=0;i nums=new ArrayList<>();
        while (!bst.isEmpty()){
            nums.add(bst.removeMax());
        }
        System.out.println(nums);
    }
}