c++ - 在二叉搜索树中查找元素仅在 true 时有效

Question

当我向我的 bst 发送查询以根据是否找到元素返回 t/f 值时，它在 true 时正常工作，但当元素不在 bst 中时会导致延迟崩溃。

我这辈子都想不通。

.cpp

bool BinarySearchTree::find(std::string title){

    if (root == NULL)
        return 0;

    return findHelper(root,title);
}

bool BinarySearchTree::findHelper(Node* current, std::string title){

    if (current->title.compare(title) == 0)
        return 1;

    if (current->title.compare(title) < 0)
        findHelper(current->left, title);
    else
        findHelper(current->right, title);

    return 0;
}

主要

if (select == 4) {

            bool x;

            string s;

            cout << "Enter Title: " << endl;

            cin.ignore();
            getline(cin, s);

            x = t.find(s);
            if (x)
                cout << "found" << endl;
            else
                cout << "unfound" << endl;

            printMenu();

            cout << "Enter Choice: ";

            cin >> select;

}

谢谢。

Answer 1

当你进行递归时，你应该返回递归调用返回的任何内容，而不是 0。只有到达树的末尾时才返回 0。

bool BinarySearchTree::findHelper(Node* current, std::string title)
{
    if (current == nullptr)
        return 0;

    if (current->title.compare(title) == 0)
        return 1;

    if (current->title.compare(title) < 0)
        return findHelper(current->left, title);
    else
        return findHelper(current->right, title);
}