Delete Node at End in DSA Rust

Mental Model

To remove the last item from a chain, find the second last item and cut its link to the last one.

Analogy: Imagine a line of people holding hands. To remove the last person, the second last person lets go of their hand, so the last person is no longer connected.

head -> 1 -> 2 -> 3 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null, delete node at end

Goal: Remove the last node (3) so the list ends at 2

Step 1: start at head node (1)

head -> [curr->1] -> 2 -> 3 -> null

Why: We begin traversal from the start to find the last node

Step 2: move curr to next node (2)

head -> 1 -> [curr->2] -> 3 -> null

Why: Keep moving forward to find the node before the last

Step 3: move curr to next node (3)

head -> 1 -> 2 -> [curr->3] -> null

Why: Reached last node, need to find previous node to remove last

Step 4: move prev to 2 and curr to 3, then set prev.next to null

head -> 1 -> [prev->2] -> null  [curr->3] (removed)

Why: Disconnect last node by making second last node point to null

Result:

head -> 1 -> 2 -> null

Annotated Code

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

#[derive(Debug)]
struct Node {
    val: i32,
    next: Option<Rc<RefCell<Node>>>,
}

impl Node {
    fn new(val: i32) -> Rc<RefCell<Node>> {
        Rc::new(RefCell::new(Node { val, next: None }))
    }
}

fn delete_node_at_end(head: &mut Option<Rc<RefCell<Node>>>) {
    if head.is_none() {
        return; // empty list, nothing to delete
    }

    let mut curr = head.as_ref().unwrap().clone();
    let mut prev: Option<Rc<RefCell<Node>>> = None;

    // Traverse until curr.next is None (last node)
    while curr.borrow().next.is_some() {
        prev = Some(curr.clone());
        let next_node = curr.borrow().next.as_ref().unwrap().clone();
        curr = next_node;
    }

    // If prev is None, list has only one node
    if let Some(prev_node) = prev {
        prev_node.borrow_mut().next = None; // remove last node
    } else {
        *head = None; // list had one node, now empty
    }
}

fn print_list(head: &Option<Rc<RefCell<Node>>>) {
    let mut curr = head.clone();
    while let Some(node) = curr {
        print!("{} -> ", node.borrow().val);
        curr = node.borrow().next.clone();
    }
    println!("null");
}

fn main() {
    // Create list 1 -> 2 -> 3 -> null
    let node1 = Node::new(1);
    let node2 = Node::new(2);
    let node3 = Node::new(3);
    node1.borrow_mut().next = Some(node2.clone());
    node2.borrow_mut().next = Some(node3.clone());

    let mut head = Some(node1);

    print_list(&head);
    delete_node_at_end(&mut head);
    print_list(&head);
}

while curr.borrow().next.is_some() {

advance curr until last node by checking next pointer

prev = Some(curr.clone());

keep track of node before last to update its next pointer

prev_node.borrow_mut().next = None;

disconnect last node by setting second last node's next to None

*head = None;

handle single node list by setting head to None

Output

1 -> 2 -> 3 -> null

1 -> 2 -> null

Complexity Analysis

Time: O(n) because we must traverse all nodes to find the last one

Space: O(1) because we only use a few pointers, no extra data structures

vs Alternative: Compared to a doubly linked list where deletion at end is O(1), singly linked list requires O(n) traversal

Edge Cases

empty list

function returns immediately, no deletion

DSA Rust

if head.is_none() { return; }

single node list

head is set to None, list becomes empty

DSA Rust

*head = None;

When to Use This Pattern

When you need to remove the last item from a singly linked list, look for traversal to the end and updating the second last node's pointer.

Common Mistakes

Mistake: Not handling the single node case, causing a panic or incorrect state
Fix: Check if prev is None after traversal and set head to None if true

Mistake: Trying to delete last node without tracking previous node, losing access to list
Fix: Keep a prev pointer to update its next to None

Summary

Deletes the last node from a singly linked list by updating the second last node's next pointer to null.

Use when you want to remove the tail element from a singly linked list.

The key is to find the second last node to safely remove the last node without losing the 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