# Doubly Linked List Delete Python Data Structures

``````def delete(lst, value):
deleted = False
if lst.is_empty():
print("List is Empty")
return deleted

if current_node.data is value:
# Point head to the next element of the first element
# Point the next element of the first element to Nobe
current_node.next_element.previous_element = None
deleted = True  # Both links have been changed.
print(str(current_node.data) + " Deleted!")
return deleted

# Traversing/Searching for node to Delete
while current_node:
if value is current_node.data:
if current_node.next_element:
# Link the next node and the previous node to each other
prev_node = current_node.previous_element
next_node = current_node.next_element
prev_node.next_element = next_node
next_node.previous_element = prev_node
# previous node pointer was maintained in Singly Linked List

else:
current_node.previous_element.next_element = None
deleted = True
break
# previousNode = tempNode was used in Singly Linked List
current_node = current_node.next_element

if deleted is False:
print(str(value) + " is not in the List!")
else:
print(str(value) + " Deleted!")
return deleted
``````

This is what the solution that is provided in the course for deleting the node from a doubly linked list. I think this solution will not work if there is only one node in the list. Before we access the next element, we should always check if the next element is not None. And then only assign the value of previous’ next_element to current. next_element as done in line 17. Same is the case if we happen to delete last element in the list. Below code is shorter and should work for all the cases.

``````def delete(lst, value):
if lst.is_empty():
print("List is empty")
return False
if current_node.data== value:
if current_node.next_element:
current_node.next_element.previous_element= None
current_node.next_element= None
return True
while current_node.next_element:
current_node= current_node.next_element
if current_node.data== value:
current_node.previous_element.next_element= current_node.next_element
if current_node.next_element:
current_node.next_element.previous_element= current_node.previous_element
current_node.previous_element= None
current_node.next_element= None
return True
return False``````