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 the implementation of method 2. The main code which creates Balanced BST is highlighted.

## C++

`#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `/* A Doubly Linked List node that` `will also be used as a tree node */` `class` `Node ` `{ ` ` ` `public` `:` ` ` `int` `data; ` ` ` ` ` `// For tree, next pointer can be` ` ` `// used as right subtree pointer ` ` ` `Node* next; ` ` ` ` ` `// For tree, prev pointer can be` ` ` `// used as left subtree pointer ` ` ` `Node* prev; ` `}; ` ` ` `// A utility function to count nodes in a Linked List ` `int` `countNodes(Node *head); ` ` ` `Node* sortedListToBSTRecur(Node **head_ref, ` `int` `n); ` ` ` `/* This function counts the number of ` `nodes in Linked List and then calls ` `sortedListToBSTRecur() to construct BST */` `Node* sortedListToBST(Node *head) ` `{ ` ` ` `/*Count the number of nodes in Linked List */` ` ` `int` `n = countNodes(head); ` ` ` ` ` `/* Construct BST */` ` ` `return` `sortedListToBSTRecur(&head, n); ` `} ` ` ` `/* The main function that constructs ` `balanced BST and returns root of it. ` `head_ref --> Pointer to pointer to` `head node of Doubly linked list ` `n --> No. of nodes in the Doubly Linked List */` `Node* sortedListToBSTRecur(Node **head_ref, ` `int` `n) ` `{ ` ` ` `/* Base Case */` ` ` `if` `(n <= 0) ` ` ` `return` `NULL; ` ` ` ` ` `/* Recursively construct the left subtree */` ` ` `Node *left = sortedListToBSTRecur(head_ref, n/2); ` ` ` ` ` `/* head_ref now refers to middle node,` ` ` `make middle node as root of BST*/` ` ` `Node *root = *head_ref; ` ` ` ` ` `// Set pointer to left subtree ` ` ` `root->prev = left; ` ` ` ` ` `/* Change head pointer of Linked List` ` ` `for parent recursive calls */` ` ` `*head_ref = (*head_ref)->next; ` ` ` ` ` `/* 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(head_ref, n-n/2-1); ` ` ` ` ` `return` `root; ` `} ` ` ` `/* UTILITY FUNCTIONS */` `/* A utility function that returns ` `count of nodes in a given Linked List */` `int` `countNodes(Node *head) ` `{ ` ` ` `int` `count = 0; ` ` ` `Node *temp = head; ` ` ` `while` `(temp) ` ` ` `{ ` ` ` `temp = temp->next; ` ` ` `count++; ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `/* Function to insert a node at ` `the beginging of the Doubly Linked List */` `void` `push(Node** head_ref, ` `int` `new_data) ` `{ ` ` ` `/* allocate node */` ` ` `Node* new_node = ` `new` `Node();` ` ` ` ` `/* put in the data */` ` ` `new_node->data = new_data; ` ` ` ` ` `/* since we are adding at the beginning, ` ` ` `prev is always NULL */` ` ` `new_node->prev = NULL; ` ` ` ` ` `/* link the old list off the new node */` ` ` `new_node->next = (*head_ref); ` ` ` ` ` `/* change prev of head node to new node */` ` ` `if` `((*head_ref) != NULL) ` ` ` `(*head_ref)->prev = new_node ; ` ` ` ` ` `/* move the head to point to the new node */` ` ` `(*head_ref) = new_node; ` `} ` ` ` `/* Function to print nodes in a given linked list */` `void` `printList(Node *node) ` `{ ` ` ` `while` `(node!=NULL) ` ` ` `{ ` ` ` `cout<<node->data<<` `" "` `; ` ` ` `node = node->next; ` ` ` `} ` `} ` ` ` `/* A utility function to print` `preorder traversal of BST */` `void` `preOrder(Node* node) ` `{ ` ` ` `if` `(node == NULL) ` ` ` `return` `; ` ` ` `cout<<node->data<<` `" "` `; ` ` ` `preOrder(node->prev); ` ` ` `preOrder(node->next); ` `} ` ` ` `/* Driver code*/` `int` `main() ` `{ ` ` ` `/* Start with the empty list */` ` ` `Node* head = NULL; ` ` ` ` ` `/* Let us create a sorted linked list to test the functions ` ` ` `Created linked list will be 7->6->5->4->3->2->1 */` ` ` `push(&head, 7); ` ` ` `push(&head, 6); ` ` ` `push(&head, 5); ` ` ` `push(&head, 4); ` ` ` `push(&head, 3); ` ` ` `push(&head, 2); ` ` ` `push(&head, 1); ` ` ` ` ` `cout<<` `"Given Linked List\n"` `; ` ` ` `printList(head); ` ` ` ` ` `/* Convert List to BST */` ` ` `Node *root = sortedListToBST(head); ` ` ` `cout<<` `"\nPreOrder Traversal of constructed BST \n "` `; ` ` ` `preOrder(root); ` ` ` ` ` `return` `0; ` `} ` ` ` `// This code is contributed by rathbhupendra` |

## 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` `int` `countNodes(` `struct` `Node *head);` ` ` `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 */` `struct` `Node* sortedListToBST(` `struct` `Node *head)` `{` ` ` `/*Count the number of nodes in Linked List */` ` ` `int` `n = countNodes(head);` ` ` ` ` `/* Construct BST */` ` ` `return` `sortedListToBSTRecur(&head, n);` `}` ` ` `/* The main function that constructs balanced BST and returns root of it.` ` ` `head_ref --> Pointer to pointer to head node of Doubly linked list` ` ` `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*/` ` ` `struct` `Node *root = *head_ref;` ` ` ` ` `// Set pointer to left subtree` ` ` `root->prev = left;` ` ` ` ` `/* Change head pointer of Linked List for parent recursive calls */` ` ` `*head_ref = (*head_ref)->next;` ` ` ` ` `/* 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(head_ref, n-n/2-1);` ` ` ` ` `return` `root;` `}` ` ` `/* UTILITY FUNCTIONS */` `/* A utility function that returns count of nodes in a given Linked List */` `int` `countNodes(` `struct` `Node *head)` `{` ` ` `int` `count = 0;` ` ` `struct` `Node *temp = head;` ` ` `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 beginning,` ` ` `prev is always NULL */` ` ` `new_node->prev = NULL;` ` ` ` ` `/* link the old list off the new node */` ` ` `new_node->next = (*head_ref);` ` ` ` ` `/* change prev of head node to new node */` ` ` `if` `((*head_ref) != NULL)` ` ` `(*head_ref)->prev = new_node ;` ` ` ` ` `/* move the head to point to the new node */` ` ` `(*head_ref) = 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()` `{` ` ` `/* Start with the empty list */` ` ` `struct` `Node* head = NULL;` ` ` ` ` `/* Let us create a sorted linked list to test the functions` ` ` `Created linked list will be 7->6->5->4->3->2->1 */` ` ` `push(&head, 7);` ` ` `push(&head, 6);` ` ` `push(&head, 5);` ` ` `push(&head, 4);` ` ` `push(&head, 3);` ` ` `push(&head, 2);` ` ` `push(&head, 1);` ` ` ` ` `printf` `(` `"Given Linked List\n"` `);` ` ` `printList(head);` ` ` ` ` `/* Convert List to BST */` ` ` `struct` `Node *root = sortedListToBST(head);` ` ` `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` `;` ` ` `}` `}` ` ` `class` `LinkedList` `{` ` ` `Node head;` ` ` ` ` `/* 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 */` ` ` `int` `n = countNodes(head);` ` ` ` ` `/* 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*/` ` ` `Node root = head;` ` ` ` ` `// Set pointer to left subtree` ` ` `root.prev = left;` ` ` ` ` `/* Change head pointer of Linked List for parent` ` ` `recursive calls */` ` ` `head = head.next;` ` ` ` ` `/* 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` ` ` `given Linked List */` ` ` `int` `countNodes(Node head)` ` ` `{` ` ` `int` `count = ` `0` `;` ` ` `Node temp = head;` ` ` `while` `(temp != ` `null` `)` ` ` `{` ` ` `temp = temp.next;` ` ` `count++;` ` ` `}` ` ` `return` `count;` ` ` `}` ` ` ` ` `/* Function to insert a node at the beginging of` ` ` `the Doubly Linked List */` ` ` `void` `push(` `int` `new_data)` ` ` `{` ` ` `/* allocate node */` ` ` `Node new_node = ` `new` `Node(new_data);` ` ` ` ` `/* since we are adding at the beginning,` ` ` `prev is always NULL */` ` ` `new_node.prev = ` `null` `;` ` ` ` ` `/* link the old list off the new node */` ` ` `new_node.next = head;` ` ` ` ` `/* change prev of head node to new node */` ` ` `if` `(head != ` `null` `)` ` ` `head.prev = new_node;` ` ` ` ` `/* move the head to point to the new node */` ` ` `head = new_node;` ` ` `}` ` ` ` ` `/* Function to print nodes in a given linked list */` ` ` `void` `printList()` ` ` `{` ` ` `Node node = head;` ` ` `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)` ` ` `{` ` ` `LinkedList llist = ` `new` `LinkedList();` ` ` ` ` `/* 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` `);` ` ` ` ` `System.out.println(` `"Given Linked List "` `);` ` ` `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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