C - Delete all nodes by key of the Doubly Linked List

In this method, all nodes in the doubly linked list with specified key (value) is deleted. For example - if the given list is 10->20->30->10->20 and all occurrences of 20 are deleted, the list becomes 10->30->10.

If the head of the doubly linked list is not null and the value stored at head is equal to key, then adjust the head as head next, delete previous head and adjust links. Keep on doing the same process until the new head becomes null or not equal to the key. After that create a temp node to traverse through the list and delete those nodes with value equal to the key and adjust links accordingly.

The function pop_all is created for this purpose. It is a 2-step process.

void pop_all(struct Node** head_ref, int key) {     
  struct Node* nodeToDelete;
  
  //1. if the head is not null and value stored at head
  //   is equal to key, keep on adjusting the head as
  //   head next, delete previous head and adjust links
  //   until new head becomes null or not equal to key
  while(*head_ref != NULL && (*head_ref)->data == key) {
    nodeToDelete = *head_ref;
    *head_ref = (*head_ref)->next;
    free(nodeToDelete);
    if(*head_ref != NULL)
      (*head_ref)->prev = NULL;
  } 

  //2. create a temp node to traverse through the
  //   list and delete nodes with value equal to 
  //   key and adjust links accordingly
  struct Node* temp = *head_ref;        
  if(temp != NULL) {
    while(temp->next != NULL) {
      if(temp->next->data == key) {
        nodeToDelete = temp->next;
        temp->next = temp->next->next;
        if(temp->next != NULL)
          temp->next->prev = temp;  
        free(nodeToDelete);
      } else {
        temp = temp->next;
      }
    }
  }
}

The below is a complete program that uses above discussed concept to delete all occurrences of the specified key (if exists) of the doubly linked list.

#include <stdio.h>
#include <stdlib.h>

//node structure
struct Node {
  int data;
  struct Node* next;
  struct Node* prev;
};

//Add new element at the end of the list
void push_back(struct Node** head_ref, int newElement) {  
  struct Node *newNode, *temp;
  newNode = (struct Node*)malloc(sizeof(struct Node)); 
  newNode->data = newElement;  
  newNode->next = NULL;
  newNode->prev = NULL;
  if(*head_ref == NULL) {
    *head_ref = newNode;
  } else {
    temp = *head_ref;
    while(temp->next != NULL) {
      temp = temp->next;
    }    
    temp->next = newNode;
    newNode->prev = temp;
  }
}

//Delete all nodes by key
void pop_all(struct Node** head_ref, int key) {     
  struct Node* nodeToDelete;
  while(*head_ref != NULL && (*head_ref)->data == key) {
    nodeToDelete = *head_ref;
    *head_ref = (*head_ref)->next;
    free(nodeToDelete);
    if(*head_ref != NULL)
      (*head_ref)->prev = NULL;
  } 

  struct Node* temp = *head_ref;        
  if(temp != NULL) {
    while(temp->next != NULL) {
      if(temp->next->data == key) {
        nodeToDelete = temp->next;
        temp->next = temp->next->next;
        if(temp->next != NULL)
          temp->next->prev = temp;  
        free(nodeToDelete);
      } else {
        temp = temp->next;
      }
    }
  }
}     

//display the content of the list
void PrintList(struct Node* head_ref) {
  struct Node* temp = head_ref;
  if(head_ref != NULL) {
    printf("The list contains: ");
    while (temp != NULL) {
      printf("%i ",temp->data);
      temp = temp->next;  
    }
    printf("\n");
  } else {
    printf("The list is empty.\n");
  }   
}

// test the code
int main() {
  struct Node* MyList = NULL;

  //Add five elements at the end of the list.
  push_back(&MyList, 10);
  push_back(&MyList, 20);
  push_back(&MyList, 30);
  push_back(&MyList, 10);
  push_back(&MyList, 20);
  PrintList(MyList);

  //Delete all occurrences of 20
  pop_all(&MyList, 20);
  PrintList(MyList);  

  return 0; 
}

The above code will give the following output:

The list contains: 10 20 30 10 20 
The list contains: 10 30 10