public string ConvertWithoutUsingAdditionDataStructure(BiNode root)
{
BiNode result = ConvertMethod2(root);
return(ConvertLinkedlistToString(result));
}
public string ConvertUsingAdditionDataStructure(BiNode root)
{
NodePair result = ConvertMethod1(root);
return(ConvertLinkedlistToString(result.Head));
}
public BiNode(int data, BiNode left, BiNode right)
{
this.Data = data;
this.Node1 = left;
this.Node2 = right;
}
public NodePair(BiNode head, BiNode tail)
{
this.Head = head;
this.Tail = tail;
}
/* This is the recursive function that converts the binary tree to linked list and returns the BiNode which is circular */
//makes the result circular
//UPDATE:
//Here is the logic
// partition 1 + root + partition 3
//(Head to tail) + root + (head to tail)
//Make it circular - connect tail of part3 to head of part1 and vice versa
//If either of the part is null
//If part1 is null - connect tail of part3 to root
//if part3 is null - connect root to part1
//dubug the code to understand
//update: do revision often
public BiNode ConvertToCircular(BiNode root)
{
if (root == null)
{
return(null);
}
//Part1 maintains the head and builds the list forward ..here it is circular
BiNode part1 = ConvertToCircular(root.Node1);
//part3 maintains the head of the second partition..and the result node is circular
BiNode part3 = ConvertToCircular(root.Node2);
if (part1 == null && part3 == null)
{
//if both part1 and part3 are null make the root circular and return
root.Node1 = root;
root.Node2 = root;
return(root);
}
//tail3 maintain the tail of the sec partition..part3.node1 will be tail bczo it is circular
BiNode tail3 = (part3 == null) ? root : part3.Node1;
/* join the left to root to form this fashion //(Head to tail) + root + (head to tail) */
if (part1 == null)
{
//here the logic is hard
//part3.node1 will be the tail and we are adding tail + root to make it circular
//no worries somewhere in the below root will be added before tail.and will make it as root + tail + root
//For eg think of this ////5////
////////////////////////////////6//
////////////////////////////////////7
//for R(5) part1 = null and part3 = 6 <-> 7 and circular
//6,7, 5
Concat(part3.Node1, root);
}
else
{
//if part1 is not null
//part1.node1 will be the tail of part1 bcoz of circular nature
//add the tail of first partition to the root
Concat(part1.Node1, root); //straight forward
}
//note not null has logic and the null code is to make circular
/* join the right to root to form this fashion //(Head to tail) + root + (head to tail) */
if (part3 == null)
{
Concat(root, part1);
}
else
{
//(Head to tail) + root + (head to tail)
Concat(root, part3);
}
/* join right to left */
///we did the circular logic for null if both are not null..make it circular here
if (part1 != null && part3 != null)
{
Concat(tail3, part1);
}
//return the head of the circular linked list
return(part1 == null ? root : part1);
}
