# In-place conversion of Sorted DLL to Balanced BST

Given a Doubly Linked List which has data members sorted in ascending order. Construct a Balanced Binary Search Tree which has same data members as the given Doubly Linked List. The tree must be constructed in-place (No new node should be allocated for tree conversion)

Examples:

```Input:  Doubly Linked List 1  2  3
Output: A Balanced BST
2
/  \
1    3

Input: Doubly Linked List 1  2 3  4 5  6  7
Output: A Balanced BST
4
/   \
2     6
/  \   / \
1   3  4   7

Input: Doubly Linked List 1  2  3  4
Output: A Balanced BST
3
/  \
2    4
/
1

Input:  Doubly Linked List 1  2  3  4  5  6
Output: A Balanced BST
4
/   \
2     6
/  \   /
1   3  5
```

The Doubly Linked List conversion is very much similar to this Singly Linked List problem and the method 1 is exactly same as the method 1 of previous post. Method 2 is also almost same. The only difference in method 2 is, instead of allocating new nodes for BST, we reuse same DLL nodes. We use prev pointer as left and next pointer as right.

Method 1 (Simple)
Following is a simple algorithm where we first find the middle node of list and make it root of the tree to be constructed.

```1) Get the Middle of the linked list and make it root.
2) Recursively do same for left half and right half.
a) Get the middle of left half and make it left child of the root
created in step 1.
b) Get the middle of right half and make it right child of the
root created in step 1.

```

Time complexity: O(nLogn) where n is the number of nodes in Linked List.

Method 2 (Tricky)
The method 1 constructs the tree from root to leaves. In this method, we construct from leaves to root. The idea is to insert nodes in BST in the same order as the appear in Doubly Linked List, so that the tree can be constructed in O(n) time complexity. We first count the number of nodes in the given Linked List. Let the count be n. After counting nodes, we take left n/2 nodes and recursively construct the left subtree. After left subtree is constructed, we assign middle node to root and link the left subtree with root. Finally, we recursively construct the right subtree and link it with root.
While constructing the BST, we also keep moving the list head pointer to next so that we have the appropriate pointer in each recursive call.
Following is C implementation of method 2. The main code which creates Balanced BST is highlighted.

## C++

```#include<stdio.h>
#include<stdlib.h>

/* A Doubly Linked List node that will also be used as a tree node */
struct Node
{
int data;

// For tree, next pointer can be used as right subtree pointer
struct Node* next;

// For tree, prev pointer can be used as left subtree pointer
struct Node* prev;
};

// A utility function to count nodes in a Linked List

struct Node* sortedListToBSTRecur(struct Node **head_ref, int n);

/* This function counts the number of nodes in Linked List and then calls
sortedListToBSTRecur() to construct BST */
{
/*Count the number of nodes in Linked List */

/* Construct BST */
}

/* The main function that constructs balanced BST and returns root of it.
n  --> No. of nodes in the Doubly Linked List */
struct Node* sortedListToBSTRecur(struct Node **head_ref, int n)
{
/* Base Case */
if (n <= 0)
return NULL;

/* Recursively construct the left subtree */
struct Node *left = sortedListToBSTRecur(head_ref, n/2);

/* head_ref now refers to middle node, make middle node as root of BST*/

// Set pointer to left subtree
root->prev = left;

/* Change head pointer of Linked List for parent recursive calls */

/* Recursively construct the right subtree and link it with root
The number of nodes in right subtree  is total nodes - nodes in
left subtree - 1 (for root) */

return root;
}

/* UTILITY FUNCTIONS */
/* A utility function that returns count of nodes in a given Linked List */
{
int count = 0;
while(temp)
{
temp = temp->next;
count++;
}
return count;
}

/* Function to insert a node at the beginging of the Doubly Linked List */
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));

/* put in the data  */
new_node->data  = new_data;

/* since we are adding at the begining,
prev is always NULL */
new_node->prev = NULL;

/* link the old list off the new node */

/* change prev of head node to new node */

/* move the head to point to the new node */
}

/* Function to print nodes in a given linked list */
void printList(struct Node *node)
{
while (node!=NULL)
{
printf("%d ", node->data);
node = node->next;
}
}

/* A utility function to print preorder traversal of BST */
void preOrder(struct Node* node)
{
if (node == NULL)
return;
printf("%d ", node->data);
preOrder(node->prev);
preOrder(node->next);
}

/* Driver program to test above functions*/
int main()
{

/* Let us create a sorted linked list to test the functions
Created linked list will be 7->6->5->4->3->2->1 */

/* Convert List to BST */
printf("\n PreOrder Traversal of constructed BST \n ");
preOrder(root);

return 0;
}
```

## Java

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

Node(int d)
{
data = d;
next = prev = null;
}
}

{

/* This function counts the number of nodes in Linked List
and then calls sortedListToBSTRecur() to construct BST */
Node sortedListToBST()
{
/*Count the number of nodes in Linked List */

/* Construct BST */
return sortedListToBSTRecur(n);
}

/* The main function that constructs balanced BST and
returns root of it.
n  --> No. of nodes in the Doubly Linked List */
Node sortedListToBSTRecur(int n)
{
/* Base Case */
if (n <= 0)
return null;

/* Recursively construct the left subtree */
Node left = sortedListToBSTRecur(n / 2);

/* head_ref now refers to middle node,
make middle node as root of BST*/

// Set pointer to left subtree
root.prev = left;

recursive calls */

/* Recursively construct the right subtree and link it
with root. The number of nodes in right subtree  is
total nodes - nodes in left subtree - 1 (for root) */
root.next = sortedListToBSTRecur(n - n / 2 - 1);

return root;
}

/* UTILITY FUNCTIONS */
/* A utility function that returns count of nodes in a
{
int count = 0;
while (temp != null)
{
temp = temp.next;
count++;
}
return count;
}

/* Function to insert a node at the beginging of
void push(int new_data)
{
/* allocate node */
Node new_node = new Node(new_data);

/* since we are adding at the begining,
prev is always NULL */
new_node.prev = null;

/* link the old list off the new node */

/* change prev of head node to new node */

/* move the head to point to the new node */
}

/* Function to print nodes in a given linked list */
void printList()
{
while (node != null)
{
System.out.print(node.data + " ");
node = node.next;
}
}

/* A utility function to print preorder traversal of BST */
void preOrder(Node node)
{
if (node == null)
return;
System.out.print(node.data + " ");
preOrder(node.prev);
preOrder(node.next);
}

/* Drier program to test above functions */
public static void main(String[] args)
{

/* Let us create a sorted linked list to test the functions
Created linked list will be 7->6->5->4->3->2->1 */
llist.push(7);
llist.push(6);
llist.push(5);
llist.push(4);
llist.push(3);
llist.push(2);
llist.push(1);

llist.printList();

/* Convert List to BST */
Node root = llist.sortedListToBST();
System.out.println("");
System.out.println("Pre-Order Traversal of constructed BST ");
llist.preOrder(root);
}
}
// This code has been contributed by Mayank Jaiswal(mayank_24)
```

Output:

```Given Linked List
1 2 3 4 5 6 7
Pre-Order Traversal of constructed BST
4 2 1 3 6 5 7
```

Time Complexity: O(n)

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