def reverse (item, tail = None):

next = item.next

item.next = tail

if next is None:

return item

else:

return reverse(next, item)

使用这样一个简单的链表实现：

class LinkedList:

def __init__ (self, value, next = None):

self.value = value

self.next = next

def __repr__ (self):

return 'LinkedList({}, {})'.format(self.value, repr(self.next))

例：

>>> a = LinkedList(1, LinkedList(2, LinkedList(3, LinkedList(4))))

>>> a

LinkedList(1, LinkedList(2, LinkedList(3, LinkedList(4, None))))

>>> b = reverse(a)

>>> b

LinkedList(4, LinkedList(3, LinkedList(2, LinkedList(1, None))))

>>> a # note that there is a new head pointer now

LinkedList(1, None)