Skip to main content

Delete a node from doubly linked list

Deletion operation in DLL is simpler when compared to SLL. Because we don't have to go in search of previous node of to-be-deleted node. 

Here is how you delete a node
  1. Link previous node of node of to-be-deleted to next node.
  2. Link next node of node of to-be-deleted to previous node.
  3. Free the memory of node of to-be-deleted
Simple, isn't it. The code can go like this.
  1. prevnode = delnode->prev;
  2. nextnode = delnode->next;
  3. prevnode->next = nextnode;
  4. nextnode->prev = prevnode;
  5. free(delnode);
And that is it. The node delnode is deleted.

But we should always consider boundary conditions. What happens if we are trying to delete the first node or last node?

If first node is to be deleted, its previous node is NULL. Hence step 3 should not be used.  And also, once head is deleted, nextnode becomes head.

Similarly if last node is to be deleted, nextnode is NULL. Hence step 4 is as strict NO NO. And we should set prevnode to tail.

After we put these things together, we have
  1. prevnode = delnode->prev;
  2. nextnode = delnode->next;
  3. if(prevnode) prevnode->next = nextnode;
  4. if(nextnode) nextnode->prev = prevnode;
  5. if(delnode==head) head = nextnode;
  6. if (delnode==tail) tail = prevnode;
  7. free(delnode); 
 Are we ready for the code now? Here it is


  1.  
 void delete_node(NODEPTR*head, NODEPTR *tail,int value)  
{
NODEPTR prevnode,nextnode;
NODEPTR delnode = find_node(*head,value);
if(delnode==NULL)
{
printf("NO such node..");
return;
}
prevnode = delnode->prev;
nextnode = delnode->next;
if(prevnode) //we have a previous node
prevnode->next = nextnode;
if(nextnode) //we have a next node
nextnode->prev = prevnode;
//if head is deleted, adjust head
if(delnode==*head)
*head = nextnode;
//if tail is deleted, adjust tail
if(delnode==*tail)
*tail = prevnode;
free(delnode);
}

You can download the complete program here.

Comments

Popular posts from this blog

Josephus problem

Question: Write a function to delete every k th node from circular linked list until only one node is left. This has a story associated with it. Flavius Josephus was Jewish Historian from 1st century. He and 40 other soldiers were trapped in a cave by Romans. They decided to kill themselves rather than surrendering to Romans. Their method was like this. All the soldiers will stand in a circle and every k th soldier will be shot dead. Josephus said to have calculated the starting point so that he would remain alive. So we have similar problem at hand. We delete every kth node in a circular list. Eventually only one node will be left. e.g. Let us say this is our list And we are deleting every third node.  We will delete 30. Then we delete 60. Next we delete 10. Next it will be 50. Next to be deleted is 20. Next 80. This continues. Implementation   We can count k-1 nodes and delete next node. This can be repeated in  a loop. What must be the termina...

Lowest common ancestor of binary search tree

Question : Write a function to print the lowest common ancestor of two nodes in a binary search tree.  Lowest common ancestor of two nodes x and y in a binary tree is the lowest node that has both x and y as descendants. Here lowest common ancestor of 1 and 7 is 3. LCA of 13 and 7 is root - 8. And LCA of 6 and 7 is 6 itself. The program to find gets complicated for an ordinary binary tree. But for a binary search tree, it is quite simple. As we see from the diagram above, the paths to 1 and 4 are common till the node 3. And at 3 they branch in different directions. So 3 is our LCA. That is lowest common ancestor is the node where the paths to x and y from root deviate. As long as they branch in same direction, we continue to traverse. When they branch in different directions, that is the lowest common ancestor. So let us write a simple algorithm, set temp=root if temp->val >x and temp->val>y temp = temp->left else if temp->val<x and ...

In order traversal of nodes in the range x to y

Question : Write a function for in-order traversal of nodes in the range x to y from a binary search tree. This is quite a simple function. As a first solution we can just traverse our binary search tree in inorder and display only the nodes which are in the range x to y. But if the current node has a value less than x, do we have to traverse its left subtree? No. Because all the nodes in left subtree will be smaller than x. Similarly if the current node has a key value more than y, we need not visit its right subtree. Now we are ready to write our algorithm.     if nd is NOT NULL  if nd->val >=x then visit all the nodes of left subtree of nd recursively display nd->val if nd->val <y then visit all the nodes of right subtree of nd recursively  That's all. We have our function ready. void in_order_middle (NODEPTR nd, int x, int y) { if (nd) { if (nd -> val >= x) in_order_middle(nd...