Delete by Value in Doubly Linked List in DSA Rust - Execution Trace

Concept Flow - Delete by Value in Doubly Linked List

Start at head

↓

Check if current node's value == target?

No→Move to next node

↓

Reached end?

↓

Adjust prev and next pointers to remove current

↓

Free current node

↓

Done or continue if multiple deletions allowed

Start from the head node, check each node's value. If it matches the target, adjust pointers of neighbors to remove it, then free the node. Continue until done.

Execution Sample

DSA Rust

fn delete_by_value(head: &mut Option<Box<Node>>, val: i32) {
    let mut current = head.as_mut();
    while let Some(node) = current {
        if node.value == val {
            // Remove node logic
            break;
        }
        current = node.next.as_mut();
    }
}

This code searches for the first node with the given value and removes it from the doubly linked list.

Execution Table

Step	Operation	Current Node Value	Pointer Changes	Visual State
1	Start at head	1	None	Head -> [1]<-> [2]<-> [3]<-> None
2	Check if current node value == 2	1	No change	Head -> [1]<-> [2]<-> [3]<-> None
3	Move to next node	2	None	Head -> [1]<-> [2]<-> [3]<-> None
4	Check if current node value == 2	2	Yes, remove node	Head -> [1]<-> [3]<-> None
5	Adjust pointers: node.prev.next = node.next, node.next.prev = node.prev	2	1.next = 3, 3.prev = 1	Head -> [1]<-> [3]<-> None
6	Free node with value 2	2	Node removed	Head -> [1]<-> [3]<-> None
7	Stop - node deleted	None	None	Head -> [1]<-> [3]<-> None

💡 Node with value 2 found and deleted; traversal stops.

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 6	Final
current	Node(1)	Node(2)	Node(2)	Node(2)	None
head	Node(1)	Node(1)	Node(1)	Node(1)	Node(1)
list size	3 nodes	3 nodes	3 nodes	2 nodes	2 nodes

Key Moments - 3 Insights

Why do we adjust both prev.next and next.prev pointers when deleting a node?

Because the list is doubly linked, both neighbors must update their pointers to skip the deleted node, as shown in step 5 of the execution_table.

What happens if the node to delete is the head node?

We must update the head pointer to the next node to keep the list accessible. This example deletes a middle node, but the same pointer adjustments apply.

Why do we stop traversal after deleting the first matching node?

Because this function deletes only the first occurrence. The exit note in the execution_table at step 7 confirms stopping after deletion.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the current node's value?

A 3

B 1

C 2

D None

💡 Hint

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

At which step do the pointers get adjusted to remove the node?

A Step 5

B Step 4

C Step 3

D Step 6

💡 Hint

Look at the 'Pointer Changes' column to find when neighbors update their links.

If the node with value 2 was not found, what would happen to 'current' after step 3?

A It would stay at Node(2)

B It would become None after reaching the end

C It would point back to head

D It would skip Node(3)

💡 Hint

Refer to the 'variable_tracker' for 'current' progression and the exit condition in the execution_table.

Concept Snapshot

Delete by Value in Doubly Linked List:
- Start at head node
- Traverse nodes checking value
- If match found, adjust prev.next and next.prev
- Remove node from memory
- Stop after first deletion
- Handles middle, head, or tail nodes

Full Transcript

To delete a node by value in a doubly linked list, start at the head and check each node's value. When the target value is found, adjust the previous node's next pointer and the next node's previous pointer to bypass the current node. Then free the current node's memory. This process stops after deleting the first matching node. The example shows deleting node with value 2 from a list of three nodes: 1, 2, 3. After deletion, the list links 1 directly to 3.

Practice

1. What is the main purpose of the delete_by_value function in a doubly linked list?

easy

A. To remove the first node that contains the given value

B. To add a new node with the given value at the end

C. To reverse the entire doubly linked list

D. To count how many nodes have the given value

Solution

1. Step 1: Understand the function's goal

   The delete_by_value function searches for a node with the specified value.

2. Step 2: Identify the action performed

   Once found, it removes that node from the doubly linked list, adjusting pointers accordingly.

3. Final Answer:

   To remove the first node that contains the given value -> Option A

4. Quick Check:

   Delete by value means remove matching node [OK]

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

Common Mistakes:

• Confusing delete with insert
• Thinking it reverses the list
• Assuming it counts nodes

2. Which Rust syntax correctly declares a mutable doubly linked list node struct with prev and next pointers?

easy

A. struct Node { value: i32, prev: Option>, next: Option> }

B. struct Node { value: i32, prev: Option<*mut Node>, next: Option<*mut Node> }

C. struct Node { value: i32, prev: Option, next: Option }

D. struct Node { value: i32, prev: Option<Node>, next: Option<Node> }

Solution

1. Step 1: Understand Rust pointer types for doubly linked lists

   Rust needs shared ownership and mutability for prev and next pointers, so Rc is used.

2. Step 2: Check options for correct pointer usage

   struct Node { value: i32, prev: Option>, next: Option> } uses Option> which allows multiple owners and interior mutability, suitable for doubly linked lists.

3. Final Answer:

   struct Node { value: i32, prev: Option>, next: Option> } -> Option A

4. Quick Check:

   Use Rc and RefCell for mutable shared nodes [OK]

Hint: Use Rc for mutable shared pointers [OK]

Common Mistakes:

• Using Box for both prev and next causes ownership issues
• Using raw pointers without unsafe code
• Using direct Node type causing infinite size

3. Given the following Rust code snippet for deleting a value from a doubly linked list, what will be the printed list after deleting value 3?

let mut list = DoublyLinkedList::new();
list.push_back(1);
list.push_back(3);
list.push_back(5);
list.delete_by_value(3);
list.print();

medium

A. 1 -> 3 -> 5 -> null

B. 1 -> 5 -> null

C. 3 -> 5 -> null

D. 1 -> null

Solution

1. Step 1: Understand initial list state

   After pushing 1, 3, 5, the list is 1 -> 3 -> 5 -> null.

2. Step 2: Delete node with value 3

   The delete_by_value(3) removes the node containing 3, linking 1 directly to 5.

3. Final Answer:

   1 -> 5 -> null -> Option B

4. Quick Check:

   Deleted 3, list is 1 -> 5 [OK]

Hint: Deleting value removes that node, links neighbors [OK]

Common Mistakes:

• Not updating prev and next pointers correctly
• Deleting wrong node or none
• Printing old list without deletion

4. Consider this Rust function snippet for deleting a node by value in a doubly linked list:

fn delete_by_value(&mut self, val: i32) {
    let mut current = self.head.clone();
    while let Some(node) = current {
        if node.borrow().value == val {
            // deletion logic here
            break;
        }
        current = node.borrow().next.clone();
    }
}

What is a common bug if the deletion logic forgets to update the prev pointer of the next node?

medium

A. The list will reverse automatically

B. The head pointer will always point to null

C. The deleted node will remain in memory forever

D. The list may become corrupted with dangling backward links

Solution

1. Step 1: Identify pointer updates needed on deletion

   When deleting a node, the previous node's next and the next node's prev pointers must be updated to bypass the deleted node.

2. Step 2: Consequence of missing prev update

   If the next node's prev pointer is not updated, it still points to the deleted node, causing backward links to be incorrect and list corruption.

3. Final Answer:

   The list may become corrupted with dangling backward links -> Option D

4. Quick Check:

   Missing prev update breaks backward links [OK]

Hint: Always update both prev and next pointers when deleting [OK]

Common Mistakes:

• Updating only next pointer
• Not breaking loop after deletion
• Forgetting to handle head or tail cases

5. You have a doubly linked list with nodes: 2 -> 4 -> 6 -> 4 -> 8 -> null. You want to delete all nodes with value 4 using a delete_by_value function that removes only the first matching node per call. How many times must you call delete_by_value(4) to remove all 4s, and what will the final list look like?

hard

A. Call once; final list: 2 -> 4 -> 6 -> 8 -> null

B. Call once; final list: 2 -> 6 -> 4 -> 8 -> null

C. Call twice; final list: 2 -> 6 -> 8 -> null

D. Call twice; final list: 2 -> 4 -> 6 -> 8 -> null

Solution

1. Step 1: Understand deletion per call

   Each call to delete_by_value(4) removes only the first node with value 4.

2. Step 2: Count how many 4s and calls needed

   There are two nodes with value 4. So, two calls are needed to remove both.

3. Step 3: Final list after deletions

   After two deletions, the list becomes 2 -> 6 -> 8 -> null.

4. Final Answer:

   Call twice; final list: 2 -> 6 -> 8 -> null -> Option C

5. Quick Check:

   Two 4s, two deletes, final list without 4s [OK]

Hint: Delete one node per call; repeat to remove all matches [OK]

Common Mistakes:

• Assuming one call removes all matching nodes
• Not updating pointers after each deletion
• Confusing node values or order