This community-built FAQ covers the “Removing the Tail” exercise from the lesson “Doubly Linked Lists: Python”.
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Pass the Technical Interview with Python
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When the list has one element, it means the head node and the tail node of the list are the same. Both methods, remove_tail() and remove_head() call each other, and it ends up in an infinite loop. Am I missing something, or this is the expected behavior?
I get a recursion error. Is there something wrong about my code? Is this the infinite loop the other poster pointed out?
class Node:
def __init__(self, value, prev_node=None, next_node=None):
self.value = value
self.prev_node = prev_node
self.next_node = next_node
def set_next_node(self, next_node):
self.next_node = next_node
def set_prev_node(self, prev_node):
self.prev_node = prev_node
def get_next_node(self):
return self.next_node
def get_prev_node(self):
return self.prev_node
def get_value(self):
return self.value
class DoublyLinkedList:
def __init__(self):
self.head_node = None
self.tail_node = None
def add_to_head(self, new_value):
new_head = Node(new_value)
current_head = self.head_node
if current_head is not None:
current_head.set_prev_node(new_head)
new_head.set_next_node(current_head)
self.head_node = new_head
if self.tail_node is None:
self.tail_node = new_head
def add_to_tail(self, new_value):
new_tail = Node(new_value)
current_tail = self.tail_node
if current_tail is not None:
current_tail.set_next_node(new_tail)
new_tail.set_prev_node(current_tail)
self.tail_node = new_tail
if self.head_node is None:
self.head_node = new_tail
def remove_head(self):
removed_head = self.head_node
new_head = removed_head.next_node
if removed_head == None:
return None
if new_head != None:
new_head.set_prev_node(None)
self.head_node = new_head
if removed_head == self.tail_node:
self.remove_tail()
return removed_head.get_value()
def remove_tail(self):
removed_tail = self.tail_node
new_tail = self.tail_node.prev_node
if removed_tail == None:
return None
if new_tail != None:
new_tail.set_next_node(None)
self.tail_node = new_tail
if removed_tail == self.head_node:
self.remove_head()
return removed_tail.get_value()
It should not end up in an infinite loop bcz if there is only one node, the tail node and head node will each be set to None over the course of remove_head() and remove_tail(), respectively.
