java如何实现一个高效的二叉搜索树（BST）？请给出时间复杂度分析。

在Java中，我们可以使用AVL树来实现一个高效的二叉搜索树。AVL树是一种自平衡的二叉搜索树，它的每个节点的两个子树的高度最多相差1，因此它的时间复杂度为O(log n)。

以下是一个简单的AVL树实现：

class Node {
   
    int key, height;
    Node left, right;

    Node(int d) {
   
        key = d;
        height = 1;
    }
}

class AVLTree {
   
    Node root;

    int height(Node N) {
   
        if (N == null)
            return 0;
        return N.height;
    }

    int max(int a, int b) {
   
        return (a > b) ? a : b;
    }

    Node rightRotate(Node y) {
   
        Node x = y.left;
        Node T2 = x.right;
        x.right = y;
        y.left = T2;
        y.height = max(height(y.left), height(y.right)) + 1;
        x.height = max(height(x.left), height(x.right)) + 1;
        return x;
    }

    Node leftRotate(Node x) {
   
        Node y = x.right;
        Node T2 = y.left;
        y.left = x;
        x.right = T2;
        x.height = max(height(x.left), height(x.right)) + 1;
        y.height = max(height(y.left), height(y.right)) + 1;
        return y;
    }

    int getBalance(Node N) {
   
        if (N == null)
            return 0;
        return height(N.left) - height(N.right);
    }

    Node insert(Node node, int key) {
   
        if (node == null)
            return (new Node(key));
        if (key < node.key)
            node.left = insert(node.left, key);
        else if (key > node.key)
            node.right = insert(node.right, key);
        else
            return node;
        node.height = 1 + max(height(node.left), height(node.right));
        int balance = getBalance(node);
        if (balance > 1 && key < node.left.key)
            return rightRotate(node);
        if (balance < -1 && key > node.right.key)
            return leftRotate(node);
        if (balance > 1 && key > node.left.key) {
   
            node.left = leftRotate(node.left);
            return rightRotate(node);
        }
        if (balance < -1 && key < node.right.key) {
   
            node.right = rightRotate(node.right);
            return leftRotate(node);
        }
        return node;
    }

    Node minValueNode(Node node) {
   
        Node current = node;
        while (current.left != null)
            current = current.left;
        return current;
    }

    Node deleteNode(Node root, int key) {
   
        if (root == null)
            return root;
        if (key < root.key)
            root.left = deleteNode(root.left, key);
        else if (key > root.key)
            root.right = deleteNode(root.right, key);
        else {
   
            if ((root.left == null) || (root.right == null)) {
   
                Node temp = null;
                if (temp == root.left)
                    temp = root.right;
                else
                    temp = root.left;
                if (temp == null) {
   
                    temp = root;
                    root = null;
                } else
                    root = temp;
            } else {
   
                Node temp = minValueNode(root.right);
                root.key = temp.key;
                root.right = deleteNode(root.right, temp.key);
            }
        }
        if (root == null)
            return root;
        root.height = max(height(root.left), height(root.right)) + 1;
        int balance = getBalance(root);
        if (balance > 1 && getBalance(root.left) >= 0)
            return rightRotate(root);
        if (balance > 1 && getBalance(root.left) < 0) {
   
            root.left = leftRotate(root.left);
            return rightRotate(root);
        }
        if (balance < -1 && getBalance(root.right) <= 0)
            return leftRotate(root);
        if (balance < -1 && getBalance(root.right) > 0) {
   
            root.right = rightRotate(root.right);
            return leftRotate(root);
        }
        return root;
    }

    boolean search(Node root, int key) {
   
        if (root == null)
            return false;
        if (root.key == key)
            return true;
        return key < root.key ? search(root.left, key) : search(root.right, key);
    }
}

        

          
        
        
        

          

          AI 代码解读
        
      
      

在这个实现中，我们使用了旋转操作来保持树的平衡。插入、删除和查找操作的时间复杂度都是O(log n)。