C++ 链表删除整个列表，而不仅仅是 1 个节点

问：

我创建了一个名为 ScanList 的链表，其中包含对象 Scan 的节点。但是，当我从链表中删除第一个扫描并关闭程序时，我在 ScanList 的析构函数处收到“访问冲突读取位置”类型的异常。

我从ScanList中删除Scan的功能如下。

bool ScanList::removeScan(int serialNumber)
{// post: IF serialNumber is present in the list 
 // THEN scan has been removed from it and true has been returned 
 // ELSE false has been returned 
    Scan* currentScan = scans;
    Scan* previousScan = scans;

    if (currentScan != NULL && currentScan->getSerialNumber() == serialNumber)
    { // if serialNumber is the first element of the linked list
        *scans = currentScan->getNext();
        delete currentScan;
        return true;
    }

    while (currentScan != NULL && currentScan->getSerialNumber() != serialNumber)
    {
        *previousScan = currentScan;
        currentScan = currentScan->getNext();
    }

    if (currentScan == NULL)
    { // serialNumber is not present in the list
        return false;
    }

    // serialNumber is present in the list and is removed
    previousScan->setNext(currentScan->getNext());
    delete currentScan;
    return true;
}

我使用复制分配运算符，但即便如此，它也会删除扫描。它确实返回正确的值，但是当我删除 currentScan 时，它也会删除 ScanList 的整个 Scan* 扫描。这是我的复制分配运算符。

Scan* Scan::operator=(const Scan* scan)
{
    // check if the Scan on the left is the same as Scan on the right
    if (this == scan)
    { // if they are the same, go out of the function and return
        return this;
    }

    this->serialNumber = scan->serialNumber;
    this->timesRecycled = scan->timesRecycled;
    this->next = scan->next;

    return this;
}

谁能告诉我我做错了什么？