def deleteNode(llist, position): if llist==None: return None else: pre=llist count=1 while pre.next is not None and count

Python

def deleteNode(llist, position): if position == 0: llist = llist.next else:
count = 1 temp = llist while temp != None and count < position: temp = temp.next count += 1

    # print(temp.data)
    temp.next = temp.next.next
    # print(temp.data)

return llist

simple and descritive python solution

def deleteNode(llist: SinglyLinkedListNode, position: int):
    if llist == None:
        return
    
    if position == 0:
        return llist.next
        
    position_counter: int = 0
    current: SinglyLinkedListNode = llist
    
    while current != None and position_counter < position - 1:
        position_counter += 1
        current = current.next

    current.next = current.next.next
    return llist

def deleteNode(head, position): if position == 0: head = head.next else: temp = head count = 1 while temp != None and count < position: temp = temp.next count += 1 temp.next = temp.next.next return head

Python

def deleteNode(llist, position):
    if position == 0:
        llist = llist.next
    else:
        previous = llist
        for _ in range(position - 1):
            previous = previous.next
        to_delete = previous.next
        following = to_delete.next
        previous.next = following
        del to_delete
    return llist