为什么链表中的节点不能负责删除自己？(C++)

我的想法是，一旦第一个节点从链表类中删除，其余的节点就会随之而来。我的实现在实践中不起作用。是否有解决方案可以逐个节点删除整个列表，而不是链表删除所有列表？我的nodeT应该有一个析构函数吗？

链表实现：

#include <iostream>
#include <cassert>
#include "nodeT.h"
class linkedListSort
{
public:
    int print();
      //Function to output the elements of the list
      //Postcondition: Elements of the list are output on the 
      //   standard output device. Member current is reset to beginning node
      //   and currentIndex is reset to 0
      //Returns:  number of items printed in the list
      //          also outputs error if number of items printed
      //          does not equal the length of the list
    void insertAt(int location, elemType& insertItem);
      //Function to insert an item in the list at the 
      //position specified by location. The item to be inserted 
      //is passed as a parameter to the function.
      //Postcondition: Starting at location, the elements of the
      //    list are shifted down, list[location] = insertItem;,
      //    and length++;. If the list is full or location is
      //    out of range, an appropriate message is displayed.
      linkedListSort(int size = 100);
      ~linkedListSort();
protected:
  //consider making const
  nodeT<elemType> *beginningNode;  // handle to the beginning of the list
  nodeT<elemType> *current;  // pointer to current node
  int currentIndex;  //int representing which node in the list current is pointing to
  int length;      //to store the length of the list
  int maxSize;     //to store the maximum size of the list
  };
template <class elemType>
linkedListSort<elemType>::linkedListSort(int size)
{
  if (size < 0)
  {
    cerr << "The array size must be positive. Creating "
      << "an array of size 100. " << endl;
    maxSize = 100;
  }
  else
    maxSize = size;
  beginningNode = NULL;
  current = NULL; // initialize to empty linked list
  length = 0;
  currentIndex = -1; // there is no item that current points to
}
template <class elemType>
linkedListSort<elemType>::~linkedListSort()
{
  delete beginningNode;  // this should delete all linked list items ( see nodeT destructor )
}

链表节点实现：

template <class elemType>
class nodeT {
public:
  nodeT(elemType& infoParam, nodeT<elemType> *linkParam);  //standard
  nodeT(elemType& infoParam);  //if unlinked node (ex. last item)
  nodeT();
  //copy constructor
  nodeT(nodeT<elemType>& node);
  ~nodeT();
  elemType *info;
  nodeT *link;
};
template<class elemType>
nodeT<elemType>::nodeT(elemType& infoParam, nodeT<elemType> *linkParam) {
  info = &infoParam;
  link = linkParam;
}
//when link is null (last item and uncircular)
template<class elemType>
nodeT<elemType>::nodeT(elemType& infoParam) {
  info = &infoParam;
  link = NULL;
}
//in case node is needed before info or link is known (default)
template<class elemType>
nodeT<elemType>::nodeT() {
  info = NULL;
  link = NULL;
}
template<class elemType>
nodeT<elemType>::nodeT(nodeT<elemType>& node) {
  info = new elemType();
  if (node.link != NULL)
    link = new nodeT();
  *info = *(node.info);  // copy by value
  if (node.link != NULL)
    *link = *(node.link);
  else
    link = NULL;
}
template<class elemType>
nodeT<elemType>::~nodeT() {
  delete info;
  if (link != NULL)
    delete link;
}

实现的最后一部分是节点析构函数。如果链接nodeT的成员类型为 nodeT则代码delete link将调用相同的析构函数，但仅在另一个实例上调用。因此，一旦第一个节点被销毁，每个节点都应该自行销毁。第一个节点在链表实现中被销毁如下：delete beginningNode beginningNode始终指向链表中的第一个节点。

我接近解决方案了吗？还是我只是掉进了一个C++不想让你掉下去的兔子洞？

实际错误与断言失败有关。然后最终我可以将其复制到我的剪贴板："第 10Ex1 章中的 0x553056E8 （msvcr120d.dll 处未处理的异常.exe：0xC0000005：访问冲突读取位置0x00000002。