实现一个简单的二分搜索树

实现一个简单的二分搜索树，有如下功能

• 获取size
• 判断是否为空
• 添加元素
• 判断是否包含指定元素
• 前序遍历
• 非递归前序遍历
• 中序遍历
• 后序遍历
• 广度优先遍历
• 查找树中的最小值
• 查找树中的最大值
• 删除最小值
• 删除最大值
• 删除指定元素

练习源码

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

/**
 * 二分搜索树
 */
public class BST<E extends Comparable<E>> {

    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 int size() {
        return size;
    }

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

    public void add(E e) {
        root = add(root, e);
    }

    /**
     * 添加元素
     *
     * @param node Node
     * @param e    要添加的元素
     * @return 添加后的Node
     */
    private Node add(Node node, E e) {
        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);
    }

    private boolean contains(Node node, E e) {
        if (node == null) {
            return false;
        }
        if (e.compareTo(node.e) == 0) {
            return true;
        }
        if (e.compareTo(node.e) < 0) {
            return contains(node.left, e);
        }
        return contains(node.right, e);
    }

    /**
     * 二叉搜索树的前序遍历
     */
    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);
    }

    /**
     * 非递归前序遍历
     */
    public void preOrderNoRecursive() {
        Stack<Node> 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);
            }
        }
    }

    /**
     * 二叉搜索树的中序遍历
     */
    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);
    }

    /**
     * 二叉搜索树的后序遍历
     */
    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);
    }

    /**
     * 二叉搜索树的广度优先遍历
     */
    public void levelOrder() {
        Queue<Node> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            Node cur = queue.remove();
            System.out.println(cur.e);
            if (cur.left != null) {
                queue.add(cur.left);
            }
            if (cur.right != null) {
                queue.add(cur.right);
            }
        }
    }

    /**
     * 查找二叉搜索树中的最小值
     *
     * @return 最小值
     */
    public E findMin() {
        if (size == 0) {
            throw new IllegalArgumentException("tree is empty");
        }
        return findMin(root).e;
    }

    private Node findMin(Node node) {
        if (node.left == null) {
            return node;
        }
        return findMin(node.left);
    }

    /**
     * 查找二叉搜索树中的最大值
     *
     * @return 最大值
     */
    public E findMax() {
        if (size == 0) {
            throw new IllegalArgumentException("tree is empty");
        }
        return findMax(root).e;
    }

    private Node findMax(Node node) {
        if (node.right == null) {
            return node;
        }
        return findMax(node.right);
    }

    /**
     * 删除最小值
     *
     * @return 删除的最小值
     */
    public E removeMin() {
        E min = findMin();
        root = removeMin(root);
        return min;
    }

    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;
    }

    /**
     * 删除最大值
     *
     * @return 删除的最大值
     */
    public E removeMax() {
        E max = findMax();
        root = removeMax(root);
        return max;
    }

    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;
    }

    /**
     * 删除指定元素
     *
     * @param e 要删除的元素
     */
    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);
            return node;
        }
        if (e.compareTo(node.e) > 0) {
            node.right = remove(node.right, e);
            return node;
        }
        if (node.left == null) {
            // 要删除的元素左节点为null
            Node rightNode = node.right;
            node.right = null;
            size--;
            return rightNode;
        }
        if (node.right == null) {
            // 要删除的元素右节点为null
            Node leftNode = node.left;
            node.left = null;
            size--;
            return leftNode;
        }

        // 待删除元素左右节点子树均不为空
        // 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
        // 用这个节点顶替待删除节点的位置
        Node successor = findMin(node.right);
        successor.right = removeMin(node.right);
        successor.left = node.left;
        node.left = node.right = null;
        return successor;
    }

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

    private void generateTreeString(Node node, int depth, StringBuilder res) {
        if (node == null) {
            res.append(generateDepthString(depth)).append("null\n");
            return;
        }
        res.append(generateDepthString(depth)).append(node.e).append("\n");
        generateTreeString(node.left, depth + 1, res);
        generateTreeString(node.right, depth + 1, res);
    }

    private String generateDepthString(int depth) {
        // JDK11 语法
        return "--".repeat(Math.max(0, depth));
    }

    public static void main(String[] args) {
        BST<Integer> tree = new BST<>();
        int[] nums = {1, 7, 3, 6, 2, 8, 11, 2};
        for (int num : nums) {
            tree.add(num);
        }
        System.out.println(tree);
    }

}