Delete Node at Beginning in DSA Rust - Execution Trace

Concept Flow - Delete Node at Beginning

Start

↓

Check if list is empty?

Yes→Nothing to delete, EXIT

No↓

Save reference to head node

↓

Move head pointer to next node

↓

Old head node is removed

↓

End

The flow checks if the list is empty, if not, it moves the head pointer to the next node, effectively deleting the first node.

Execution Sample

DSA Rust

struct Node {
    data: i32,
    next: Option<Box<Node>>,
}

fn delete_at_beginning(head: &mut Option<Box<Node>>) {
    if head.is_some() {
        *head = head.as_mut().unwrap().next.take();
    }
}

This Rust function deletes the first node of a singly linked list by moving the head pointer to the next node.

Execution Table

Step	head before	Condition head.is_some()	Action	head after	Output
1	Some(Node(data=10, next=Some(Node(data=20, next=None))))	True	Move head to next node	Some(Node(data=20, next=None))	List after deletion: 20 -> null
2	Some(Node(data=20, next=None))	True	Move head to next node	None	List after deletion: null
3	None	False	No deletion, list empty	None	List is empty, nothing to delete

💡 head becomes None, list is empty, deletion stops

Variable Tracker

Variable	Start	After 1	After 2	Final
head	Some(Node(data=10))	Some(Node(data=20))	None	None

Key Moments - 2 Insights

Why do we check if head.is_some() before deleting?

Because if head is None (empty list), there is no node to delete. See execution_table step 3 where condition is False and no deletion happens.

What happens to the old head node after deletion?

The old head node is no longer referenced by head and is dropped by Rust's memory management. This is shown in execution_table step 1 where head moves from Node(10) to Node(20).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of head after step 1?

A None

B Some(Node(data=10, next=...))

C Some(Node(data=20, next=None))

D Some(Node(data=30, next=None))

💡 Hint

Check the 'head after' column in execution_table row for step 1.

At which step does the list become empty?

A Step 2

B Step 1

C Step 3

D Never

💡 Hint

Look at the 'head after' column and see when it becomes None.

If the list starts empty (head = None), what happens when delete_at_beginning is called?

A head becomes Some(Node)

B head remains None, no deletion

C Function crashes

D head points to a new node

💡 Hint

See execution_table step 3 where condition is False and head stays None.

Concept Snapshot

Delete Node at Beginning:
- Check if list is empty (head is None)
- If not empty, move head pointer to next node
- Old head node is removed automatically
- Safe in Rust due to ownership and Option type
- Result: first node deleted, list updated

Full Transcript

This concept shows how to delete the first node of a singly linked list in Rust. We start by checking if the list is empty using head.is_some(). If the list is not empty, we move the head pointer to the next node by taking ownership of the next node. This effectively removes the first node from the list. If the list is empty, no deletion happens. The execution table traces these steps with the head pointer values before and after deletion. The variable tracker shows how head changes from pointing to the first node, then the second, and finally None when the list is empty. Key moments clarify why the check for empty list is necessary and what happens to the old head node. The visual quiz tests understanding of head pointer changes and behavior on empty lists.

Practice

1. What does deleting a node at the beginning of a singly linked list do to the list?

easy

A. Deletes all nodes in the list

B. Removes the last node and updates the tail

C. Removes the first node and updates the head to the next node

D. Removes a node from the middle of the list

Solution

1. Step 1: Understand the head pointer role

   The head points to the first node in the list, so deleting the first node means changing the head.

2. Step 2: Update the head to the next node

   After removing the first node, the head should point to the second node, effectively removing the first node from the list.

3. Final Answer:

   Removes the first node and updates the head to the next node -> Option C

4. Quick Check:

   Delete at beginning = remove first node [OK]

Hint: Deleting at start means moving head to next node [OK]

Common Mistakes:

• Trying to remove the last node instead of the first
• Not updating the head pointer after deletion
• Deleting all nodes instead of just one

2. Which Rust code snippet correctly deletes the first node of a singly linked list stored in a variable head?

easy

A. head = head.take().and_then(|node| node.next);

B. head = head.unwrap().next;

C. head = head.next;

D. head.next = head;

Solution

1. Step 1: Understand Rust Option handling for linked lists

   In Rust, linked lists often use Option. To delete the first node, we take the head and replace it with its next node safely.

2. Step 2: Use take() and and_then() to update head

   head.take() removes the current head, and and_then(|node| node.next) sets head to the next node, deleting the first.

3. Final Answer:

   head = head.take().and_then(|node| node.next); -> Option A

4. Quick Check:

   Use take() and and_then() to delete first node [OK]

Hint: Use take() and and_then() to update head safely [OK]

Common Mistakes:

• Trying to assign head.next directly without handling Option
• Using unwrap() which can panic if head is None
• Assigning head.next to head causing infinite loop

3. What will be the output linked list after deleting the first node from this Rust linked list?

let mut head = Some(Box::new(Node { value: 10, next: Some(Box::new(Node { value: 20, next: Some(Box::new(Node { value: 30, next: None })) })) }));
head = head.take().and_then(|node| node.next);

Print the list values in order.

medium

A. []

B. [10, 20, 30]

C. [30]

D. [20, 30]

Solution

1. Step 1: Initial list values

   The list starts as 10 -> 20 -> 30 -> None.

2. Step 2: Delete the first node

   Using head.take().and_then(|node| node.next) removes the node with value 10 and sets head to node with value 20.

3. Final Answer:

   [20, 30] -> Option D

4. Quick Check:

   After deletion, list starts at 20 [OK]

Hint: Deleting first node removes head value [OK]

Common Mistakes:

• Forgetting to update head, so list stays same
• Assuming deletion removes last node
• Confusing node values order after deletion

4. Identify the error in this Rust code that tries to delete the first node of a linked list:

fn delete_first(mut head: Option) -> Option {
    head = head.unwrap().next;
    head
}

medium

A. Using unwrap() without checking if head is None causes panic

B. The function does not return the updated head

C. head.next is assigned incorrectly

D. The function should take &mut head instead of ownership

Solution

1. Step 1: Check unwrap() usage

   Calling unwrap() on an Option that might be None will cause the program to panic at runtime.

2. Step 2: Safe deletion requires handling None

   Before unwrap(), code should check if head is Some or use methods like take() and and_then() to avoid panic.

3. Final Answer:

   Using unwrap() without checking if head is None causes panic -> Option A

4. Quick Check:

   unwrap() on None causes panic [OK]

Hint: Never unwrap() without checking Option first [OK]

Common Mistakes:

• Ignoring possibility of empty list (None)
• Not returning updated head properly
• Confusing ownership and borrowing in function

5. Given a Rust singly linked list where each node contains an integer, write the correct way to delete the first node and return the updated list head. Which code snippet correctly handles the edge case when the list is empty?

hard

A. fn delete_first(head: Option) -> Option {
head.unwrap().next
}

B. fn delete_first(mut head: Option) -> Option {
head = head.take().and_then(|node| node.next);
head
}

C. fn delete_first(head: Option) -> Option {
if head.is_some() {
head.unwrap()
} else {
None
}
}

D. fn delete_first(head: Option) -> Option {
head.next
}

Solution

1. Step 1: Handle Option safely with take() and and_then()

   Using head.take().and_then(|node| node.next) safely removes the first node and returns the next node or None if empty.

2. Step 2: Return updated head

   Assign the result back to head and return it, ensuring the function works for empty and non-empty lists.

3. Final Answer:

   fn delete_first(mut head: Option) -> Option { head = head.take().and_then(|node| node.next); head } -> Option B

4. Quick Check:

   Use take() and and_then() for safe deletion [OK]

Hint: Use take() and and_then() to handle empty list safely [OK]

Common Mistakes:

• Using unwrap() without checking for None
• Accessing head.next directly on Option type
• Not returning the updated head after deletion