Skip to main content

Convert a BST to a DLL

I came across this question in the net.

Given a binary search tree, convert this into a sorted doubly linked list. Do not create any nodes, but just modify the links.

Just like a DLL, binary tree node also has two links. But they are named left and right. What if we assume left is previous link and right is next link? OK, we don't even have to modify the structure of the nodes.

Next important part is "sorted". We need a sorted DLL. Don't you get the idea now. Which traversal visits the nodes in ascending order or gives sorted output?

In order traversal! That's right. So we need to visit the nodes in in-order traversal, store previous node in a static variable, and link this previous node to the next node visited using previous (left) and next(right) links.

Now how do we determine the head of DLL? It should be the tree-minimum. So we can find tree minimum and set this as head. Or we can use in order traversal and set the node which has no previous node visited as head.


void convert_to_dll(NODEPTR nd,NODEPTR*headptr)
{
   static NODEPTR prevnode;
   if(nd!=NULL)
    {       
 convert_to_dll(nd->left,headptr);
        if(prevnode!=NULL){
           prevnode->right = nd;
           nd->left = prevnode;
        }
        else{
           *headptr = nd;
        }
        prevnode = nd; 
        convert_to_dll(nd->right,headptr);
     }
}

Now let us look at the complete program.


#include<stdio.h>
#include<stdlib.h>
struct node
{
   int val;
   struct node *left;
   struct node *right;
};
typedef struct node *NODEPTR;

NODEPTR create_node(int num)
{
     NODEPTR temp = (NODEPTR)malloc(sizeof(struct node));
     temp->val = num;
     temp->left = NULL;
     temp->right = NULL;
     return temp;
} 

NODEPTR insert_node(NODEPTR nd,NODEPTR newnode)
{
    if(nd==NULL)
       return newnode;/* newnode becomes root of tree*/
    if(newnode->val > nd->val)
        nd->right = insert_node(nd->right,newnode);
    else if(newnode->val <  nd->val)
        nd->left = insert_node(nd->left,newnode); 
    return nd;   
}

void convert_to_dll(NODEPTR nd,NODEPTR*headptr)
{
   static NODEPTR prevnode;
   if(nd!=NULL)
    {       
 convert_to_dll(nd->left,headptr);
        if(prevnode!=NULL){
           prevnode->right = nd;
           nd->left = prevnode;
        }
        else{
           *headptr = nd;
        }
        prevnode = nd; 
        convert_to_dll(nd->right,headptr);
     }
}
 
void print_dll(NODEPTR nd)
{
   NODEPTR temp = nd;
    while(nd)
      {
        printf("%d---->",nd->val);
        temp = nd; nd = nd->right;
       }
     printf("\nnow in reverse");
     nd = temp;
     while(nd)
      {
        printf("%d---->",nd->val);
        temp = nd; nd = nd->left;
       }
} 
int main()
{
       NODEPTR root=NULL,delnode,head; 
       int n;
       do
       {
           NODEPTR newnode;
           printf("Enter value of node(-1 to exit):");
           scanf("%d",&n);
           if(n!=-1)
            {  
               newnode = create_node(n);
               root = insert_node(root,newnode);
             }
       } while (n!=-1);
       
       printf("\nInorder traversal\n");
       head = NULL;
       convert_to_dll(root,&head);
       print_dll(head); 
       return 0;
}

The driver program traverses the DLL in forward as well as reverse.

Labels:

Comments

Post a Comment

Popular posts from this blog

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...
Post a Comment
Read more

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...
Post a Comment
Read more

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 ...
Post a Comment
Read more