Convert a Sorted Doubly Linked List to Balanced BST.

Objective: Given a sorted doubly linked list, convert it into Balanced binary search tree

Input: A Doubly Linked List

Example:

Approach:

Get the size of the Doubly Linked list.
Take left n/2 nodes and recursively construct left subtree
Make the middle node as root and assign the left subtree( constructed in step 2) to root’s left.
Recursively construct right subtree and link it to the the right of root made in step 3.
See picture below

Complete Code:

	public class DLLToBST {
	public static Node head = null;
	public static Node tail = null;
	public static int size = 0;
	public Node root;

	public void add(int data) {
	Node n = new Node(data);
	if (head == null) {
	head = n;
	tail = n;
	} else {
	head.prev = n;
	n.next = head;
	head = n;
	}
	size++;
	}

	public Node dLLtoBST(int size) {
	if (size <= 0) {
	return null;
	}
	Node left = dLLtoBST(size / 2);
	Node root = head;
	root.prev = left;
	head = head.next;
	root.next = dLLtoBST(size–(size / 2)–1);
	return root;
	}

	public void inOrder(Node root) {
	if (root != null) {
	inOrder(root.prev);
	System.out.print(" " + root.data);
	inOrder(root.next);
	}
	}

	public void printDLL(Node head) {
	Node curr = head;
	while (curr != null) {
	System.out.print(" " + curr.data);
	curr = curr.next;
	}
	System.out.println();
	}

	public static void main(String args[]) {
	DLLToBST r = new DLLToBST();
	r.add(9);
	r.add(8);
	r.add(7);
	r.add(6);
	r.add(5);
	r.add(4);
	r.add(3);
	r.add(2);
	r.add(1);
	Node h = head;
	System.out.println("DLL is : ");
	r.printDLL(h);
	Node x = r.dLLtoBST(size);
	System.out.println("Inorder traversal of contructed BST");
	r.inOrder(x);
	}
	}

	class Node {
	int data;
	Node next;
	Node prev;

	public Node(int data) {
	this.data = data;
	this.next = null;
	this.prev = null;
	}
	}

Output :

DLL is :
1 2 3 4 5 6 7 8 9
Inorder traversal of contructed BST
1 2 3 4 5 6 7 8 9