Delete Node at End in DSA Rust - Execution Trace

Concept Flow - Delete Node at End

Start at head

↓

Is list empty?

Yes→Stop, nothing to delete

No↓

Is only one node?

Yes→Delete head, set head to None

No↓

Traverse to second last node

↓

Set second last node's next to None

↓

Delete last node

↓

End

Start from the head, check if list is empty or has one node, otherwise traverse to second last node and remove the last node by updating pointers.

Execution Sample

DSA Rust

fn delete_at_end(head: &mut Option<Box<Node>>) {
    if head.is_none() {
        return;
    }
    if head.as_ref().unwrap().next.is_none() {
        *head = None;
        return;
    }
    let mut current = head.as_mut().unwrap();
    while current.next.as_ref().unwrap().next.is_some() {
        current = current.next.as_mut().unwrap();
    }
    current.next = None;
}

Deletes the last node from a singly linked list by traversing to the second last node and removing the last.

Execution Table

Step	Action	Current Node Value	Next Node Value	List State
1	Check if head is None	N/A	N/A	List: 1 -> 2 -> 3 -> null
2	Check if only one node	1	2	List: 1 -> 2 -> 3 -> null
3	Set current to head	1	2	List: 1 -> 2 -> 3 -> null
4	Traverse to second last node	2	3	List: 1 -> 2 -> 3 -> null
5	Set current.next to None (delete last node)	2	None	List: 1 -> 2 -> null
6	End	2	None	List after deletion: 1 -> 2 -> null

💡 Last node deleted by setting second last node's next to None

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	Final
head	1 -> 2 -> 3 -> null	1 -> 2 -> 3 -> null	1 -> 2 -> 3 -> null	1 -> 2 -> null	1 -> 2 -> null
current	None	Node(1)	Node(2)	Node(2)	Node(2)

Key Moments - 2 Insights

Why do we check if the list has only one node separately?

Because if there is only one node, deleting it means setting head to None. This check is shown in execution_table row 2.

Why do we stop traversal at the second last node, not the last?

We need to update the second last node's next pointer to None to remove the last node. The traversal stops at the second last node as shown in execution_table row 4.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'current' after step 4?

A Node with value 2

B Node with value 3

C Node with value 1

D None

💡 Hint

Check the 'Current Node Value' column at step 4 in execution_table

At which step does the last node get deleted?

A Step 3

B Step 4

C Step 5

D Step 6

💡 Hint

Look at the 'Action' column describing deletion in execution_table

If the list had only one node, what would happen at step 2?

A The function would continue traversing

B The function would return after setting head to None

C The function would crash

D The function would delete the second node

💡 Hint

Refer to key_moments explanation about single node deletion and execution_table row 2

Concept Snapshot

Delete Node at End in Singly Linked List:
- Check if list is empty: do nothing
- If only one node: set head to None
- Else traverse to second last node
- Set second last node's next to None
- Last node is removed
- Time complexity: O(n)

Full Transcript

To delete the last node in a singly linked list, first check if the list is empty. If it is, do nothing. If the list has only one node, delete it by setting the head to None. Otherwise, start from the head and move through the list until you reach the second last node. Then, set its next pointer to None, effectively removing the last node. This process takes linear time because you may need to traverse the entire list.

Practice

1. What does the operation Delete Node at End do in a linked list?

easy

A. Adds a new node at the end of the linked list

B. Removes the last node from the linked list

C. Removes the first node from the linked list

D. Searches for a node at the end of the linked list

Solution

1. Step 1: Understand the linked list structure

   A linked list is a chain of nodes where each node points to the next one.

2. Step 2: Identify the effect of deleting the last node

   Deleting the last node means removing the node that has no next node, effectively shortening the list by one node at the end.

3. Final Answer:

   Removes the last node from the linked list -> Option B

4. Quick Check:

   Delete Node at End = Remove last node [OK]

Hint: Deleting at end always removes the last node [OK]

Common Mistakes:

• Confusing delete at end with delete at start
• Thinking it adds a node instead of removing
• Assuming it searches without deleting

2. Which of the following Rust code snippets correctly deletes the last node of a singly linked list?

easy

A. fn delete_end(head: &mut Option) { if head.is_none() { return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; }

B. fn delete_end(head: &mut Option) { *head = None; }

C. fn delete_end(head: &mut Option) { let mut current = head.as_mut().unwrap(); current.next = None; }

D. fn delete_end(head: &mut Option) { while head.is_some() { head = head.unwrap().next; } }

Solution

1. Step 1: Analyze the code logic for deleting last node

   fn delete_end(head: &mut Option) { if head.is_none() { return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; } traverses the list until the second last node and sets its next to None, removing the last node.

2. Step 2: Check other options for correctness

   fn delete_end(head: &mut Option) { *head = None; } deletes the entire list, not just last node. fn delete_end(head: &mut Option) { let mut current = head.as_mut().unwrap(); current.next = None; } incorrectly sets next of head only. fn delete_end(head: &mut Option) { while head.is_some() { head = head.unwrap().next; } } tries to reassign head incorrectly causing error.

3. Final Answer:

   Option A code correctly deletes the last node -> Option A

4. Quick Check:

   Traverse to second last node and set next to None [OK]

Hint: Traverse to second last node, then set next to None [OK]

Common Mistakes:

• Setting head to None deletes whole list
• Not checking for empty or single node list
• Incorrectly reassigning head pointer

3. Given the Rust linked list code below, what will be the printed output after deleting the last node?

struct Node { val: i32, next: Option }
fn print_list(head: &Option) {
  let mut current = head;
  while let Some(node) = current {
    print!("{} -> ", node.val);
    current = &node.next;
  }
  println!("null");
}
fn main() {
  let mut head = Some(Box::new(Node { val: 1, next: Some(Box::new(Node { val: 2, next: Some(Box::new(Node { val: 3, next: None })) })) }));
  delete_end(&mut head);
  print_list(&head);
}
fn delete_end(head: &mut Option) {
  if head.is_none() { return; }
  if head.as_ref().unwrap().next.is_none() { *head = None; return; }
  let mut current = head.as_mut().unwrap();
  while current.next.as_ref().unwrap().next.is_some() {
    current = current.next.as_mut().unwrap();
  }
  current.next = None;
}

medium

A. null

B. 1 -> 2 -> 3 -> null

C. 1 -> 2 -> null

D. 1 -> null

Solution

1. Step 1: Understand initial list and delete_end effect

   The list starts as 1 -> 2 -> 3 -> null. delete_end removes last node (3).

2. Step 2: Trace print_list after deletion

   After deletion, list is 1 -> 2 -> null, so print_list outputs "1 -> 2 -> null".

3. Final Answer:

   1 -> 2 -> null -> Option C

4. Quick Check:

   Last node removed, list ends at 2 [OK]

Hint: Deleting last node removes final value from print output [OK]

Common Mistakes:

• Expecting original list unchanged
• Confusing print output format
• Not handling single node list case

4. The following Rust function to delete the last node of a linked list causes a runtime panic. What is the error?

fn delete_end(head: &mut Option) {
  let mut current = head.as_mut().unwrap();
  while current.next.is_some() {
    current = current.next.as_mut().unwrap();
  }
  current.next = None;
}

medium

A. It panics because it sets current to None instead of current.next

B. It panics because it modifies head directly without checking

C. It panics because it never stops the loop

D. It panics because it tries to unwrap None when list is empty

Solution

1. Step 1: Locate the unwrap without check

   let mut current = head.as_mut().unwrap(); executes without prior if head.is_none() { return; }, causing panic on empty list when as_mut() returns None and unwrap panics.

2. Step 2: Verify other options

   The code does not modify head directly. The while loop terminates when current.next.is_none(). Setting current.next = None after reaching the last node causes no panic.

3. Final Answer:

   It panics because it tries to unwrap None when list is empty -> Option D

4. Quick Check:

   head.as_mut().unwrap() without empty check [OK]

Hint: Check head.is_none() before any unwrap on head [OK]

Common Mistakes:

• No check for empty list before unwrap
• Not checking for empty or single node list
• Looping until current.next is None instead of second last

5. You have a singly linked list with nodes containing values [10, 20, 30, 40]. Write a Rust function to delete the last node and return the updated list head. Which of the following implementations correctly handles the edge case when the list has only one node?

hard

A. fn delete_end(head: &mut Option) { if head.is_none() { return; } if head.as_ref().unwrap().next.is_none() { *head = None; return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; }

B. fn delete_end(head: &mut Option) { let mut current = head.as_mut().unwrap(); while current.next.is_some() { current = current.next.as_mut().unwrap(); } current = None; }

C. fn delete_end(head: &mut Option) { if head.is_none() { return; } let mut current = head.as_mut().unwrap(); while current.next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; }

D. fn delete_end(head: &mut Option) { *head = None; }

Solution

1. Step 1: Check for empty and single node list

   fn delete_end(head: &mut Option) { if head.is_none() { return; } if head.as_ref().unwrap().next.is_none() { *head = None; return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; } correctly checks if list is empty or has only one node, setting head to None in that case.

2. Step 2: Traverse to second last node and delete last

   fn delete_end(head: &mut Option) { if head.is_none() { return; } if head.as_ref().unwrap().next.is_none() { *head = None; return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; } then traverses to second last node and sets its next to None, deleting last node properly.

3. Final Answer:

   Option A correctly handles all cases including single node list -> Option A

4. Quick Check:

   Check empty/single node, then delete last node [OK]

Hint: Always check if list has one node before deleting last [OK]

Common Mistakes:

• Not handling single node list causing panic
• Setting current = None instead of current.next = None
• Deleting entire list instead of last node