CSE240 Doubly Linked ListsBY GARRETT GUTIERREZ

What is a doubly linked list?A doubly linked list is a list of elements.Each element points to the next element in

the list as well as the previous element in the list.

Why use a doubly linked list over a single linked list?Doubly linked lists allow easy traversal forward

and backward through the list.

How to get started? Create either a struct or an object to represent a node in

the linked list. It has a reference to the previous and next items in the

linked list. Use a tail pointer, in addition to a head pointer.struct node{

int id;char* name; // A pointer, not an arraystruct node* previous; // pointer to previous nodestruct node* next; // pointer to next node in list

};

Q1. What is the size of the node?.(A) 4 bytes(B) 16 bytes(C) more than 16 bytes?(D) The size is unknown at

compilation.Think: How to obtain memory for name?

Insertion: Case by Case Inserting an element into a doubly linked list

is very similar to inserting an element into a singly linked list.

Additionally, just make the previous pointer point to the previous item in the linked list.

There are three cases for sorted insert:1. Add at the beginning2. Add at an arbitrary location3. Add to the end

Case 0: The linked list is empty

Make the new node the head

Set previous and next to NULL of head

Set tail to head

head NULL

NULL

head nameidprevious next

if (head == NULL){

head = newNode;head->next = NULL;head->previous = NULL;tail = head;return;

}tail

NULL

Q2. When is the tail pointer equals to the head pointer? Select all that apply.(A) When the list is empty(B) When there is one node only(C) When there are more than one node(D) They never equal.

Case 1: Add element to beginning

Set newNode->next to head->nextSet head->previous to newNodeSet head to the newNodeMake the newNode previous NULL

head

NULLif (strcmp(newNode->name, head->name) < 0){newNode->next = head;head->previous = newNode;head = newNode;head->previous = NULL;return;}

nameidprevious next

nameidprevious next

nameidprevious next

head

NULL

newNode The old headNULL

NULL

Q3. Which node’s previous pointer is NULL?(A) The first node(B) The second node(C) The last node(D) No node.

Case 2: Add at an arbitrary locationUse a iterator pointer and a follower

pointer.follower always is one element behind

iteratorstruct node* iterator = head;struct node* follower = iterator;

Case 2: Add to an arbitrary location

Set newNode->next to iteratorSet iterator->previous to newNodeSet the newNode->previous to followerSet follower->next to newNode

head nameidprevious next

nameidprevious next

NULL

nameidprevious nextnewNode

follower iterator

NULL

Case 3: Add to the endGo through the entire list until iterator is

NULL

Set follower->next to newNodeSet newNode->previous to followerSet newNode->next to NULLSet the tail to newNode

head nameidprevious next

nameidprevious next

NULL

follower iterator

nameidprevious next

tail

NULL

Q4. Is case 3 included in case 2?(A) Yes, case 3 can be simply

removed(B) No. case 3 is not included in

case 2. Explain your answer

Putting it Togetherif (head == NULL) { // Case 0head = newNode;head->next = NULL;head->previous = NULL;tail = head;return;}else { Case 1 if (strcmp(newNode->name, head->name) < 0) {

newNode->next = head;head->previous = newNode;head = newNode;head->previous = NULL;return;

} }};

iterator = head;follower = iterator;while (iterator != NULL) { // Case 2 if (strcmp(newNode->name, iterator->name) < 0) {

newNode->next = iterator;iterator->previous = newNode;newNode->previous = follower;follower->next = newNode;return;

} follower = iterator; iterator = iterator->next;}follower->next = newNode; // Case 3newNode->previous = follower;newNode->next = NULL;tail = newNode;return;

Searching Doubly Linked List: Front

1. Start at the head.2. Go until the iterator is

NULL3. Compare the value of the

iterator with a value.4. If found, return the

element.5. Else, set iterator to next.6. If not found, return NULL.

struct node* iterator = head;while (iterator != NULL){

if (strcmp(iterator->name, token) == 0)

{return iterator;

}iterator = iterator->next;

}printf("The element does not exist.\n\n");return iterator;

Searching Doubly Linked List: Back1. Start at the tail.2. Go until the iterator is NULL3. Compare the value of the

iterator with a value.4. If found, return the element.5. Else, set iterator to previous.6. If not found, return NULL.

struct node* iterator = tail;while (iterator != NULL){

if (strcmp(iterator->name, token) == 0)

{return iterator;

}iterator = iterator->previous;

}printf("The element does not exist.\n\n");return iterator;

Q5. How many lines of code are different when searching forwards and backwards?(A) 0 (B) 1 (C) 2 (D) 4Which lines of code?

Deleting an ElementCase 1: List only has one item

Delete heap memory for nameDelete heap memory for headSet head to NULLSet tail to NULL

NULL

head nameidprevious next

if (head == toDelete && head->next == NULL){

free(head->name);free(head);head = NULL;tail = NULL;return;

}

tail

NULL

Q6. Advance Question: Why do call free twice?free(head->name);free(head);To be discussed in memory management section later.

Continue only if time permit

Deleting an ElementCase 2: Remove head only

Set head to the next item in listDelete memory for nameDelete memory of nodeMake new head->previous NULL

else if (head == toDelete){

head =head ->next;free(toDelete->name);free(toDelete);toDelete = NULL;head->previous =

NULL;return;

}

head nameidprevious next

nameidprevious next

NULL

Old head New headNULL NULL

Case 3: Delete at arbitrary location

Set follower->next to item after iteratorSet item after iterator’s previous to followerDelete memory for nameDelete memory for nodeSet toDelete and iterator to NULL

head nameidprevious next

nameidprevious next

NULLnameidprevious next

toDeletefollower iterator

NULL

Case 3: Delete at the endGo through the entire list until iterator ==

tail

Delete memory for nameDelete memory for nodeSet tail to the followerSet follower->next to NULL

head nameidprevious next

nameidprevious next

follower iterator

nameidprevious next

tail

NULLNULL

tail

if (iterator == tail){

free(iterator->name);free(iterator);tail = follower;follower->next = NULL;return;

}

Putting it all togetherif (head == toDelete && head->next == NULL){

free(toDelete->name);free(toDelete);head = NULL;tail = NULL;return;

}else if (head == toDelete){

head = head->next;free(toDelete->name);free(toDelete);toDelete = NULL;head->previous = NULL;return;

}

while (iterator != NULL){

if (iterator == tail){

free(iterator->name);free(iterator);tail = follower;follower->next = NULL;return;

}else if (strcmp(iterator->name, toDelete->name) == 0){

follower->next = iterator->next;iterator->next->previous = follower;free(iterator->name);free(iterator);iterator = NULL;toDelete = NULL;return;

}follower = iterator;iterator = iterator->next;

}

Printing Doubly Linked List: Front

1. Start at the head.2. Go until the iterator is

NULL3. Print the values of the

node.4. Set iterator to next.

struct node* iterator = head;

while (iterator != NULL){

printf("Name: %s\n”, iterator->name); printf(“ID: %d\n", iterator->id); iterator = iterator->next;

}

Printing Doubly Linked List: Back

1. Start at the tail.2. Go until the iterator is

NULL3. Print the values of the

node.4. Set iterator to previous.

struct node* iterator = tail;

while (iterator != NULL){

printf("Name: %s\n”, iterator->name); printf(“ID: %d\n", iterator->id); iterator = iterator->previous;

}