用C++解决一个问题，问题如下，设计一个算法，它从二叉检索树中删除一个人工输入的结点，并仍然保持二叉检索树的特性不变。

好的，这里是一个简单的C++代码实现，可以让用户自己输入要删除的值：

#include <iostream>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode* deleteNode(TreeNode* root, int key) {
    if (root == NULL) return root;
    if (key < root->val) {
        root->left = deleteNode(root->left, key);
    } else if (key > root->val) {
        root->right = deleteNode(root->right, key);
    } else {
        if (root->left == NULL) {
            TreeNode* temp = root->right;
            delete root;
            return temp;
        } else if (root->right == NULL) {
            TreeNode* temp = root->left;
            delete root;
            return temp;
        } else {
            TreeNode* temp = root->right;
            while (temp->left != NULL) {
                temp = temp->left;
            }
            root->val = temp->val;
            root->right = deleteNode(root->right, temp->val);
        }
    }
    return root;
}

void inorder(TreeNode* root) {
    if (root == NULL) return;
    inorder(root->left);
    cout << root->val << " ";
    inorder(root->right);
}

int main() {
    TreeNode* root = new TreeNode(5);
    root->left = new TreeNode(3);
    root->right = new TreeNode(6);
    root->left->left = new TreeNode(2);
    root->left->right = new TreeNode(4);
    root->right->right = new TreeNode(7);

    cout << "Before deletion: ";
    inorder(root);
    cout << endl;

    int key;
    cout << "Enter the value of node to be deleted: ";
    cin >> key;

    root = deleteNode(root, key);

    cout << "After deletion: ";
    inorder(root);
    cout << endl;

    return 0;
}

该算法的时间复杂度为O(h)，其中h是二叉搜索树的高度。在最坏的情况下，h可以达到O(n)，其中n是树的结点数，因此，该算法的最坏时间复杂度为O(n)。