Delete Node by Value in DSA Python - Execution Trace

Concept Flow - Delete Node by Value

Start at head

↓

Check if head is null

Yes→List empty, stop

No↓

Check if head value == target

↓

Delete head

↓

Check next node value == target

↓

Delete next node

↓

Stop or continue

Start from the head node, check if it matches the value to delete. If yes, remove it. Otherwise, move through the list to find and delete the node with the target value.

Execution Sample

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def delete_node(head, target):
    if head is None:
        return None
    if head.val == target:
        return head.next
    current = head
    while current.next:
        if current.next.val == target:
            current.next = current.next.next
            break
        current = current.next
    return head

Deletes the first node with the given target value from a singly linked list.

Execution Table

Step	Current Node Value	Check Condition	Action	List State
1	Head (1)	Is head None? No	Check if head.val == 3	1 -> 2 -> 3 -> 4 -> null
2	Head (1)	head.val == 3? No	Set current = head	1 -> 2 -> 3 -> 4 -> null
3	Current (1)	current.next exists? Yes	Check if current.next.val == 3	1 -> 2 -> 3 -> 4 -> null
4	Current.next (2)	2 == 3? No	Move current to current.next	1 -> 2 -> 3 -> 4 -> null
5	Current (2)	current.next exists? Yes	Check if current.next.val == 3	1 -> 2 -> 3 -> 4 -> null
6	Current.next (3)	3 == 3? Yes	Delete node with value 3 by linking 2 -> 4	1 -> 2 -> 4 -> null
7	Current (2)	Stop traversal	Return head	1 -> 2 -> 4 -> null

💡 Node with value 3 found and deleted; traversal stops.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	Final
head	Node(1)	Node(1)	Node(1)	Node(1)	Node(1)
current	None	Node(1)	Node(2)	Node(2)	Node(2)
current.next	Node(2)	Node(2)	Node(3)	Node(4)	Node(4)

Key Moments - 3 Insights

Why do we check if the head node has the target value before traversing?

Because if the head node itself contains the target value, we must delete it by returning head.next. This is shown in execution_table row 1 and 2 where we check and delete the head if needed.

Why do we check current.next.val instead of current.val during traversal?

We check current.next.val because to delete a node, we need to change the 'next' pointer of the previous node. This is shown in execution_table rows 3 to 6 where current.next is inspected and updated.

What happens if the target value is not found in the list?

The loop finishes without deleting any node, and the original list is returned unchanged. This is implied by the loop condition and no break in the traversal.

Visual Quiz - 3 Questions

Look at the execution_table at step 6, what action is performed?

A Move current to next node

B Delete node with value 3 by linking 2 -> 4

C Check if head is None

D Return head

💡 Hint

Refer to execution_table row 6 under 'Action' column.

At which step does the traversal stop after deleting the node?

A Step 4

B Step 5

C Step 7

D Step 3

💡 Hint

Look at execution_table row 7 where traversal stops and head is returned.

If the target value was at the head node, what would be the new head after deletion?

A The node after head

B The original head node

C None

D The last node

💡 Hint

Check execution_table row 1 and 2 where head.val is checked and head.next is returned if matched.

Concept Snapshot

Delete Node by Value in Linked List:
- Check if list is empty (head is None).
- If head has target value, delete by returning head.next.
- Else, traverse list checking current.next for target.
- If found, link current.next to current.next.next to delete.
- Return head after deletion or if not found.

Full Transcript

This visual trace shows how to delete a node by value in a singly linked list. We start by checking if the list is empty. If not, we check if the head node contains the target value. If yes, we delete the head by returning the next node. Otherwise, we traverse the list, checking the next node's value at each step. When we find the target, we delete it by linking the current node to the node after the target. The traversal stops after deletion. If the target is not found, the list remains unchanged. Variables head and current track our position. This step-by-step helps understand pointer changes during deletion.

Practice

1. What does the operation Delete Node by Value do in a linked list?

easy

A. Removes the first node that contains the given value

B. Adds a new node with the given value at the end

C. Finds the node with the given value without removing it

D. Reverses the linked list

Solution

1. Step 1: Understand the operation

   Deleting a node by value means removing the first node that matches the given value from the linked list.

2. Step 2: Compare options

   Only Removes the first node that contains the given value describes removing the first node with the given value. Others describe different operations.

3. Final Answer:

   Removes the first node that contains the given value -> Option A

4. Quick Check:

   Delete Node by Value = Remove first matching node [OK]

Hint: Delete node by value removes first matching node [OK]

Common Mistakes:

• Confusing deletion by value with adding a node
• Thinking it deletes all nodes with the value
• Mixing deletion by value with reversing the list

2. Which of the following is the correct Python syntax to delete a node by value from a singly linked list?

easy

A. def delete_node(head, value):
current = head
while current:
if current.val == value:
current = current.next
break
current = current.next

B. def delete_node(head, value):
prev = None
current = head
while current:
if current.val == value:
if prev:
prev.next = current.next
else:
head = current.next
break
prev = current
current = current.next
return head

C. def delete_node(head, value):
for node in head:
if node.val == value:
head.remove(node)
break
return head

D. def delete_node(head, value):
head = [x for x in head if x != value]
return head

Solution

1. Step 1: Analyze the deletion logic

   Deleting a node requires tracking the previous node to update its next pointer. def delete_node(head, value): prev = None current = head while current: if current.val == value: if prev: prev.next = current.next else: head = current.next break prev = current current = current.next return head correctly uses prev and current pointers and updates links properly.

2. Step 2: Check other options

   def delete_node(head, value): current = head while current: if current.val == value: current = current.next break current = current.next only moves current pointer without changing links. def delete_node(head, value): for node in head: if node.val == value: head.remove(node) break return head uses invalid iteration on linked list nodes. def delete_node(head, value): head = [x for x in head if x != value] return head treats linked list as a list, which is incorrect.

3. Final Answer:

   Option B code correctly deletes node by value in a singly linked list -> Option B

4. Quick Check:

   Proper pointer update = def delete_node(head, value): prev = None current = head while current: if current.val == value: if prev: prev.next = current.next else: head = current.next break prev = current current = current.next return head [OK]

Hint: Use prev and current pointers to update links [OK]

Common Mistakes:

• Not updating previous node's next pointer
• Trying to iterate linked list like a Python list
• Not returning updated head after deletion

3. What will be the output linked list after deleting value 3 from the list: 1 -> 3 -> 5 -> 3 -> 7 -> null?

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def delete_node(head, value):
    prev = None
    current = head
    while current:
        if current.val == value:
            if prev:
                prev.next = current.next
            else:
                head = current.next
            break
        prev = current
        current = current.next
    return head

# Create list
head = Node(1)
head.next = Node(3)
head.next.next = Node(5)
head.next.next.next = Node(3)
head.next.next.next.next = Node(7)

head = delete_node(head, 3)

# Print list
current = head
result = []
while current:
    result.append(current.val)
    current = current.next
print(result)

medium

A. [1, 5, 3, 7]

B. [1, 3, 5, 7]

C. [3, 5, 3, 7]

D. [1, 5, 7]

Solution

1. Step 1: Identify the first node with value 3

   The list is 1 -> 3 -> 5 -> 3 -> 7. The first node with value 3 is the second node.

2. Step 2: Delete the first 3 and link previous node to next

   After deletion, the list becomes 1 -> 5 -> 3 -> 7.

3. Final Answer:

   [1, 5, 3, 7] -> Option A

4. Quick Check:

   First 3 removed, rest unchanged = [1, 5, 3, 7] [OK]

Hint: Delete only first matching node, keep rest intact [OK]

Common Mistakes:

• Deleting all nodes with value 3 instead of first only
• Not updating head when first node matches
• Skipping nodes after deletion

4. The following code tries to delete a node by value from a linked list but causes an error. What is the error?

def delete_node(head, value):
    current = head
    while current.next:
        if current.next.val == value:
            current.next = current.next.next
            break
        current = current.next
    return head

medium

A. Fails when the value is not found

B. Fails when the list is empty

C. Fails when the node to delete is the last node

D. Fails when the node to delete is the head node

Solution

1. Step 1: Analyze deletion logic

   The code checks current.next.val to find the node to delete, so it cannot delete the head node if it matches the value.

2. Step 2: Check edge cases

   If the head node has the value, the code never deletes it because it only looks at current.next.

3. Final Answer:

   Fails when the node to delete is the head node -> Option D

4. Quick Check:

   Head node deletion not handled = Fails when the node to delete is the head node [OK]

Hint: Check if head node matches before loop [OK]

Common Mistakes:

• Not handling deletion of head node separately
• Assuming current.next always exists without check
• Ignoring empty list case

5. You have a singly linked list: 10 -> 20 -> 30 -> 40 -> 50 -> null. You want to delete the node with value 10 and then insert a new node with value 15 after the node with value 20. What will be the final linked list? Choose the correct sequence of operations and final list. Options: A) Delete 10, insert 15 after 20: 20 -> 15 -> 30 -> 40 -> 50 B) Delete 10, insert 15 after 20: 10 -> 20 -> 15 -> 30 -> 40 -> 50 C) Delete 10, insert 15 after 20: 20 -> 30 -> 15 -> 40 -> 50 D) Delete 10, insert 15 after 20: 15 -> 20 -> 30 -> 40 -> 50

hard

A. 20 -> 30 -> 15 -> 40 -> 50

B. 10 -> 20 -> 15 -> 30 -> 40 -> 50

C. 20 -> 15 -> 30 -> 40 -> 50

D. 15 -> 20 -> 30 -> 40 -> 50

Solution

1. Step 1: Delete node with value 10

   Since 10 is the head, deleting it makes 20 the new head. List becomes 20 -> 30 -> 40 -> 50.

2. Step 2: Insert node with value 15 after node with value 20

   Find node 20, insert 15 after it. List becomes 20 -> 15 -> 30 -> 40 -> 50.

3. Final Answer:

   20 -> 15 -> 30 -> 40 -> 50 -> Option C

4. Quick Check:

   Delete head 10, insert 15 after 20 = 20 -> 15 -> 30 -> 40 -> 50 [OK]

Hint: Delete head carefully, then insert after target node [OK]

Common Mistakes:

• Not updating head after deleting first node
• Inserting new node in wrong position
• Confusing node values and positions