在双向链表中使用的节点类与单链表中使用的节点类相比，应有何不

解：每一个结点包括数据域和指向链表中其它结点的指针（即下一个结点的地址），单链表中使用的节点类中每个结点中只有一个指向后继结点指针；在双向链表中使用的节点类中有两个用于连接其它结点的指针，一个指向前趋结点（称前趋指针），另一个指向后继结点（称后继指针）。双向链表中使用的节点类可如下定义：

//dnode.h
#ifndef DOUBLY_LINKED_NODE_CLASS
#define DOUBLY_LINKED_NODE_CLASS
template <class T>
class DNode
{
private:
circular links to the left and right
DNode<T> *left;
DNode<T> *right;
public:
//data is public
T data;
// constructors
DNode(void);
DNode (const T& item);
// list modification methods
void InsertRight(DNode<T> *p);
void InsertLeft(DNode<T> *p);
DNode<T> *DeleteNode(void);
//obtain address of the next node to the left or right 
DNode<T> *NextNodeRight(void) const;
DNode<T> *NextNodeLeft(void) const; 
};
//constructor that creates an empty list and
//leaves the data uninitialized. use for header 
template <class T>
DNode<T>::DNode(void)
{
//initialize the node so it points to itself
left = right = this;
}
//constructor that creates an empty list and initializes data 
template <class T>
DNode<T>::DNode(const T& item)
{
//set node to point to itself and initialize data
left = right = this; 
data = item;
}
//insert a node p to the right of current node 
template <class T>
void DNode<T>::InsertRight(DNode<T> *p)
{
//link p to its successor on the right
p->right = right;
right->left = p;
//link p to the current node on its left 
p->left = this;
right = p;
}
insert a node p to the left of current node 
template <class T>
void DNode<T>::InsertLeft(DNode<T> *p)
{
//link p to its successor on the left
p->left = left;
left->right = p;
// link p to the current node on its right
p->right = this;
left = p;
}
//unlink the current node from the list and return its address
template <class T>
DNode<T> *DNode<T>::DeleteNode(void)
{
//node to the left must be linked to current node's right
left->right = right;
//node to the right must be linked to current node's left
right->left = left;
//return the address of the current node
return this;
}
// return pointer to the next node on the right
template <class T>
DNode<T> *DNode<T>::NextNodeRight(void) const
{
return right;
}
// return pointer to the next node on the left
template <class T>
DNode<T> *DNode<T>::NextNodeLeft(void) const
{
return left;
}
#endif // DOUBLY_LINKED_NODE_CLASS