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.
Now let us look at the complete program.
The driver program traverses the DLL in forward as well as reverse.
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.
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