Stack pop using linked list - Python - python-3.x

Question

I am trying to pop an item off a stack (using a linked list as oppose to an array). I first created a LinkedList class with 3 nodes with the values [1,2,3]. So I would like pop off the last node (node_C, value=3), therefore I expect to see the values [1,2]. Instead nothing prints out.
class LinkedList:
def __init__(self, value):
self.value = value
self.next = None
node_A = LinkedList(1)
node_B = LinkedList(2)
node_C = LinkedList(3)
node_A.next = node_B
node_B.next = node_C
def pop(head):
current_node = head
while current_node.next:
if current_node.next == None:
break
else:
current_node = current_node.next
del current_node
return node_A.value, node_B.value, node_C.value
try:
print(pop(node_A))
except NameError:
pass
How can I rewrite this to achieve my desired results (i.e. show values 1,2 .. with 3 popped off)?

Answer 1

The del current node and return node_A.value, node_B.value, node_C.value
commands should belong to the pop function, so they should be intended. But anyway the del current node doesnt work for me. Instead you could write current_node.value = None but then you still return all 3 node values so the result would be 1,2,None.
I would rather write the pop function inside the class and add another printlist function to the class as well. The pop function just removes the last element from the list (changes the next attribute to None for the 2nd last element in the list) and doesn't print or return anything. The printlist function iterates through the list and prints out all elements (while there are next elements). Here is my code:
class LinkedList:
def __init__(self, value):
self.value = value
self.next = None
def pop(self):
current_node = self
while current_node.next:
if current_node.next.next == None:
current_node.next = None
else:
current_node = current_node.next
def printlist(self):
current_node = self
lst = [current_node.value]
while current_node.next:
current_node = current_node.next
lst.append(current_node.value)
print lst
node_A = LinkedList(1)
node_B = LinkedList(2)
node_C = LinkedList(3)
node_A.next = node_B
node_B.next = node_C
try:
node_A.pop()
node_A.printlist()
except NameError:
pass
If I run this, the result is [1,2]. If I remove the node_A.pop() I get [1,2,3]. If I write another node_A.pop() then result is [1]

Answer 2

I guess I have found the issue with your logic, so based on the code provided it seems that pop function doesn't return anything, may be it's just formatting or something else.
But here is the correct version of your code, where I just delete the last node in the pop method and I call another method called listValues which returns me with the node values that exist in the linked list after pop
Look at the below implementation for a clearer view.
class LinkedList:
def __init__(self, value):
self.value = value
self.next = None
node_A = LinkedList(1)
node_B = LinkedList(2)
node_C = LinkedList(3)
node_A.next = node_B
node_B.next = node_C
def pop(head):
current_node = head
while current_node.next:
if current_node.next == None:
del current_node
break
else:
current_node = current_node.next
def listValues(head):
values = []
current_node = head
while current_node.next:
values.append(current_node.value)
current_node = current_node.next
return values
try:
pop(node_A)
print(listValues(node_A))
except NameError:
pass
Hope this helps!