递归反转自定义链表时，head.next\u节点突然变为非

Question

我正在尝试使用递归实现反向链表：

class node:
    def __init__(self, data=None):
        self.data=data
        self.next_node=None

class linked_list:
    def __init__(self):
        self.head=node()

    def push(self, data):
        new_node=node(data)
        cur_node=self.head
        while(cur_node.next_node!=None):
            cur_node=cur_node.next_node
        cur_node.next_node=new_node



    def display(self):
        elems=[]
        cur_node=self.head
        print('Display:')
        # print(cur_node)
        # print(cur_node.next_node)
        # print(cur_node.data)
        while(cur_node.next_node!=None):
            cur_node=cur_node.next_node
            elems.append(cur_node.data)
        print(elems)


    def lenth(self):
        i=0
        cur_node=self.head
        while(cur_node.next_node!=None):
            last_node=cur_node
            cur_node=cur_node.next_node
            i+=1
        print(i)


    def reversell_rec(self, node):
        # print("Recursive")
        cur_node = node
        # print(cur_node)
        # print(cur_node.next_node)
        # print(cur_node.data)
        if (cur_node.next_node == None):
            self.head.next_node = cur_node
            return
        self.reversell_rec(cur_node.next_node)
        temp=cur_node.next_node
        temp.next_node = node
        node.next_node = None



ll=linked_list()
ll.push(1)
ll.push(2)
ll.display()
ll.reversell_rec(ll.head)
ll.display()

我得到输出：

Display:  #Display before recursion
 [1, 2]
Display:  #Display after recursion
 [] 

我试着用不同的方法打印出来，但不知怎么的self.head.next\节点即使我将最后一个节点分配给self.head.next\节点到最后一个节点。是什么导致了这种变化？你能帮忙吗？你知道吗

编辑：

def reversell_rec(self, node):
    # print("Recursive")
    cur_node = node
    # print(cur_node)
    # print(cur_node.next_node)
    # print(cur_node.data)
    if (cur_node.next_node == None):
        self.head.next_node = cur_node
        return
    self.reversell_rec(cur_node.next_node)
    if(cur_node!=self.head):
        temp = cur_node.next_node
        temp.next_node = cur_node
        cur_node.next_node = None '   

Answer 1

这看起来是可变的。如果你想这样做的话，基本的想法是通过向后积累来构造一个反向。你知道吗

def reversell_rec(self):
    def rev_node(n, acc):
        if n is None: return acc
        nxt = n.next_node
        n.next_node = acc
        return rev_node(nxt, n)

    self.head = rev_node(self.head, None)

但老实说，如果你在一个循环中做的话，如果它是可变的，那就容易多了。你知道吗