Delete Node at End in DSA Rust - Build from Scratch

Delete Node at End in a Singly Linked List

📖 Scenario: Imagine you have a chain of boxes connected one after another. Each box holds a number. You want to remove the last box from the chain.

🎯 Goal: You will build a simple singly linked list in Rust and write code to delete the last node from it.

📋 What You'll Learn

Create a singly linked list with 3 nodes containing values 10, 20, and 30

Add a helper variable to track the head of the list

Write code to delete the last node from the list

Print the list after deletion showing nodes connected by ' -> ' and ending with 'null'

💡 Why This Matters

🌍 Real World

Linked lists are used in many programs to manage collections of items where you need to add or remove items efficiently.

💼 Career

Understanding linked lists and how to manipulate them is a fundamental skill for software developers, especially when working with low-level data structures or memory management.

1

Create a singly linked list with 3 nodes

Create a struct called Node with a field value of type i32 and a field next which is an Option holding a Box<Node>. Then create three nodes with values 10, 20, and 30 linked together. Finally, create a variable called head that points to the first node.

DSA Rust

// Define the Node struct and create the linked list with 3 nodes
// Your code here

Hint

Use Option> for the next field to allow linking nodes.

2

Create a mutable variable to hold the head node

Change the variable head to be mutable by declaring it with let mut head so you can modify the list later.

DSA Rust

struct Node {
    value: i32,
    next: Option>,
}

fn main() {
    // Make head mutable
    let mut head = Node { value: 10, next: Some(Box::new(Node { value: 20, next: Some(Box::new(Node { value: 30, next: None })) })) };
    // Your code here
}

Hint

Use let mut head to allow changing the list later.

3

Delete the last node from the linked list

Write code to delete the last node from the linked list starting at head. Use a mutable reference current to traverse the list until you find the node whose next points to the last node. Then set that node's next to None to remove the last node.

DSA Rust

struct Node {
    value: i32,
    next: Option>,
}

fn main() {
    let node3 = Node { value: 30, next: None };
    let node2 = Node { value: 20, next: Some(Box::new(node3)) };
    let mut head = Node { value: 10, next: Some(Box::new(node2)) };

    // Traverse to delete last node
    // Your code here
}

Hint

Use a while let loop with mutable references to find the second last node.

4

Print the linked list after deleting the last node

Write a loop to print the values of the linked list starting from head. Print each node's value followed by -> . After the last node, print null.

DSA Rust

struct Node {
    value: i32,
    next: Option>,
}

fn main() {
    let node3 = Node { value: 30, next: None };
    let node2 = Node { value: 20, next: Some(Box::new(node3)) };
    let mut head = Node { value: 10, next: Some(Box::new(node2)) };

    let mut current = &mut head;
    while let Some(ref mut next_node) = current.next {
        if next_node.next.is_none() {
            current.next = None;
            break;
        }
        current = current.next.as_mut().unwrap();
    }

    // Print the list
    // Your code here
}

Hint

Use a loop with a reference to print each node's value followed by ' -> ', then print 'null' at the end.

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