### Deleting a node from a linked list is straightforward but there are a few cases we need to account for:

- 1. The list is empty
- 2. The node to remove is the only node in the linked list
- 3. We are removing the head node
- 4. We are removing the tail node
- 5. The node to remove is somewhere in between the head and tail node
- 6. The item to remove does not exist in the linked list.

## Step 1: If the list is empty then print a message and return.

```
def delete(self, key):
# If the list is empty
if self.head is None:
print('Deletion Error: The list is empty.')
return
```

## Step 2: If the head holds the key, delete the head by assigning head to the next node of head.

```
# If the key is in head
if self.head.data == key:
self.head = self.head.next
return
```

## Step 3: Find the first occurrence of the key in the linked list.

```
# Find position of first occurrence of the key
current = self.head
while current:
if current.data == key:
break
previous = current
current = current.next
```

## If the key was found then point the previous node to the next node of the key. Otherwise print an error message.

```
# If the key was not found
if current is None:
print('Deletion Error: Key not found.')
else:
previous.next = current.next
```

## Step 4: Check sample test cases for the code.

## The complete code is given below:

```
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def printList(self):
temp = self.head
if not temp:
print('List is empty.')
return
else:
print('Start:', end=' ')
while temp:
print(temp.data, end=' -> ')
temp = temp.next
print('end.')
def insert(self, data):
new_node = Node(data)
# If the linked list is empty
if self.head is None:
self.head = new_node
# If the data is smaller than the head
elif self.head.data >= new_node.data:
new_node.next = self.head
self.head = new_node
else:
# Locate the node before the insertion point
current = self.head
while current.next and new_node.data > current.next.data:
current = current.next
# Insertion
new_node.next = current.next
current.next = new_node
def delete(self, key):
# If the list is empty
if self.head is None:
print('Deletion Error: The list is empty.')
return
# If the key is in head
if self.head.data == key:
self.head = self.head.next
return
# Find position of first occurrence of the key
current = self.head
while current:
if current.data == key:
break
previous = current
current = current.next
# If the key was not found
if current is None:
print('Deletion Error: Key not found.')
else:
previous.next = current.next
if __name__=='__main__':
# Create an object
LL = LinkedList()
print('')
# Insert some nodes
LL.insert(10)
LL.insert(12)
LL.insert(8)
LL.insert(11)
LL.insert(10)
LL.printList()
LL.delete(7)
LL.delete(8)
LL.delete(13)
LL.printList()
```

## Thank you for reading the article. Please leave you feedback. ðŸ˜„

References:

- https://www.geeksforgeeks.org/linked-list-set-3-deleting-node/
- CLRS: Introduction to Algorithms [book]

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