Related Articles

# Sorted Linked List to Balanced BST

• Difficulty Level : Hard
• Last Updated : 02 Aug, 2021

Given a Singly Linked List which has data members sorted in ascending order. Construct a Balanced Binary Search Tree which has same data members as the given Linked List.
Examples:

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

Output: A Balanced BST
4
/   \
2     6
/  \   / \
1   3  5   7

Output: A Balanced BST
3
/  \
2    4
/
1

Output: A Balanced BST
4
/   \
2     6
/  \   /
1   3  5   ```

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

```1) Get the Middle of the linked list and make it root.
2) Recursively do same for the left half and right half.
a) Get the middle of the left half and make it left child of the root
created in step 1.
b) Get the middle of right half and make it the 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)
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 they appear in 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 allocate memory for 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 implementation of method 2. The main code which creates Balanced BST is highlighted.

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `/* Link list node */``class` `LNode``{``    ``public``:``    ``int` `data;``    ``LNode* next;``};` `/* A Binary Tree node */``class` `TNode``{``    ``public``:``    ``int` `data;``    ``TNode* left;``    ``TNode* right;``};` `TNode* newNode(``int` `data);``int` `countLNodes(LNode *head);``TNode* sortedListToBSTRecur(LNode **head_ref, ``int` `n);`  `/* This function counts the number of``nodes in Linked List and then calls``sortedListToBSTRecur() to construct BST */``TNode* sortedListToBST(LNode *head)``{``    ``/*Count the number of nodes in Linked List */``    ``int` `n = countLNodes(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 linked list n --> No.``of nodes in Linked List */``TNode* sortedListToBSTRecur(LNode **head_ref, ``int` `n)``{``    ``/* Base Case */``    ``if` `(n <= 0)``        ``return` `NULL;` `    ``/* Recursively construct the left subtree */``    ``TNode *left = sortedListToBSTRecur(head_ref, n/2);` `    ``/* Allocate memory for root, and``    ``link the above constructed left``    ``subtree with root */``    ``TNode *root = newNode((*head_ref)->data);``    ``root->left = 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) which is n-n/2-1*/``    ``root->right = 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` `countLNodes(LNode *head)``{``    ``int` `count = 0;``    ``LNode *temp = head;``    ``while``(temp)``    ``{``        ``temp = temp->next;``        ``count++;``    ``}``    ``return` `count;``}` `/* Function to insert a node``at the beginning of the linked list */``void` `push(LNode** head_ref, ``int` `new_data)``{``    ``/* allocate node */``    ``LNode* new_node = ``new` `LNode();``    ` `    ``/* put in the data */``    ``new_node->data = new_data;` `    ``/* link the old list off the new node */``    ``new_node->next = (*head_ref);` `    ``/* move the head to point to the new node */``    ``(*head_ref) = new_node;``}` `/* Function to print nodes in a given linked list */``void` `printList(LNode *node)``{``    ``while``(node!=NULL)``    ``{``        ``cout << node->data << ``" "``;``        ``node = node->next;``    ``}``}` `/* Helper function that allocates a new node with the``given data and NULL left and right pointers. */``TNode* newNode(``int` `data)``{``    ``TNode* node = ``new` `TNode();``    ``node->data = data;``    ``node->left = NULL;``    ``node->right = NULL;` `    ``return` `node;``}` `/* A utility function to``print preorder traversal of BST */``void` `preOrder(TNode* node)``{``    ``if` `(node == NULL)``        ``return``;``    ``cout<data<<``" "``;``    ``preOrder(node->left);``    ``preOrder(node->right);``}` `/* Driver code*/``int` `main()``{``    ``/* Start with the empty list */``    ``LNode* head = NULL;` `    ``/* Let us create a sorted linked list to test the functions``    ``Created linked list will be 1->2->3->4->5->6->7 */``    ``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 "``;``    ``printList(head);` `    ``/* Convert List to BST */``    ``TNode *root = sortedListToBST(head);``    ``cout<<``"\nPreOrder Traversal of constructed BST "``;``    ``preOrder(root);` `    ``return` `0;``}` `// This code is contributed by rathbhupendra`

## C

 `#include``#include` `/* Link list node */``struct` `LNode``{``    ``int` `data;``    ``struct` `LNode* next;``};` `/* A Binary Tree node */``struct` `TNode``{``    ``int` `data;``    ``struct` `TNode* left;``    ``struct` `TNode* right;``};` `struct` `TNode* newNode(``int` `data);``int` `countLNodes(``struct` `LNode *head);``struct` `TNode* sortedListToBSTRecur(``struct` `LNode **head_ref, ``int` `n);`  `/* This function counts the number of nodes in Linked List and then calls``   ``sortedListToBSTRecur() to construct BST */``struct` `TNode* sortedListToBST(``struct` `LNode *head)``{``    ``/*Count the number of nodes in Linked List */``    ``int` `n = countLNodes(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 linked list``       ``n  --> No. of nodes in Linked List */``struct` `TNode* sortedListToBSTRecur(``struct` `LNode **head_ref, ``int` `n)``{``    ``/* Base Case */``    ``if` `(n <= 0)``        ``return` `NULL;` `    ``/* Recursively construct the left subtree */``    ``struct` `TNode *left = sortedListToBSTRecur(head_ref, n/2);` `    ``/* Allocate memory for root, and link the above constructed left``       ``subtree with root */``    ``struct` `TNode *root = newNode((*head_ref)->data);``    ``root->left = 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) which is n-n/2-1*/``    ``root->right = 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` `countLNodes(``struct` `LNode *head)``{``    ``int` `count = 0;``    ``struct` `LNode *temp = head;``    ``while``(temp)``    ``{``        ``temp = temp->next;``        ``count++;``    ``}``    ``return` `count;``}` `/* Function to insert a node at the beginning of the linked list */``void` `push(``struct` `LNode** head_ref, ``int` `new_data)``{``    ``/* allocate node */``    ``struct` `LNode* new_node =``        ``(``struct` `LNode*) ``malloc``(``sizeof``(``struct` `LNode));``    ``/* put in the data  */``    ``new_node->data  = new_data;` `    ``/* link the old list off the new node */``    ``new_node->next = (*head_ref);` `    ``/* 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` `LNode *node)``{``    ``while``(node!=NULL)``    ``{``        ``printf``(``"%d "``, node->data);``        ``node = node->next;``    ``}``}` `/* Helper function that allocates a new node with the``   ``given data and NULL left and right pointers. */``struct` `TNode* newNode(``int` `data)``{``    ``struct` `TNode* node = (``struct` `TNode*)``                         ``malloc``(``sizeof``(``struct` `TNode));``    ``node->data = data;``    ``node->left = NULL;``    ``node->right = NULL;` `    ``return` `node;``}` `/* A utility function to print preorder traversal of BST */``void` `preOrder(``struct` `TNode* node)``{``    ``if` `(node == NULL)``        ``return``;``    ``printf``(``"%d "``, node->data);``    ``preOrder(node->left);``    ``preOrder(node->right);``}` `/* Driver program to test above functions*/``int` `main()``{``    ``/* Start with the empty list */``    ``struct` `LNode* head = NULL;` `    ``/* Let us create a sorted linked list to test the functions``     ``Created linked list will be 1->2->3->4->5->6->7 */``    ``push(&head, 7);``    ``push(&head, 6);``    ``push(&head, 5);``    ``push(&head, 4);``    ``push(&head, 3);``    ``push(&head, 2);``    ``push(&head, 1);` `    ``printf``(``"\n Given Linked List "``);``    ``printList(head);` `    ``/* Convert List to BST */``    ``struct` `TNode *root = sortedListToBST(head);``    ``printf``(``"\n PreOrder Traversal of constructed BST "``);``    ``preOrder(root);` `    ``return` `0;``}`

## Java

 `class` `LinkedList {` `    ``/* head node of link list */``    ``static` `LNode head;``    ` `    ``/* Link list Node */``    ``class` `LNode``    ``{``        ``int` `data;``        ``LNode next, prev;` `        ``LNode(``int` `d)``        ``{``            ``data = d;``            ``next = prev = ``null``;``        ``}``    ``}``    ` `    ``/* A Binary Tree Node */``    ``class` `TNode``    ``{``        ``int` `data;``        ``TNode left, right;` `        ``TNode(``int` `d)``        ``{``            ``data = d;``            ``left = right = ``null``;``        ``}``    ``}` `    ``/* This function counts the number of nodes in Linked List``       ``and then calls sortedListToBSTRecur() to construct BST */``    ``TNode 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 */``    ``TNode sortedListToBSTRecur(``int` `n)``    ``{``        ``/* Base Case */``        ``if` `(n <= ``0``)``            ``return` `null``;` `        ``/* Recursively construct the left subtree */``        ``TNode left = sortedListToBSTRecur(n / ``2``);` `        ``/* head_ref now refers to middle node,``           ``make middle node as root of BST*/``        ``TNode root = ``new` `TNode(head.data);` `        ``// Set pointer to left subtree``        ``root.left = 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.right = sortedListToBSTRecur(n - n / ``2` `- ``1``);` `        ``return` `root;``    ``}` `    ``/* UTILITY FUNCTIONS */``    ``/* A utility function that returns count of nodes in a``       ``given Linked List */``    ``int` `countNodes(LNode head)``    ``{``        ``int` `count = ``0``;``        ``LNode temp = head;``        ``while` `(temp != ``null``)``        ``{``            ``temp = temp.next;``            ``count++;``        ``}``        ``return` `count;``    ``}` `    ``/* Function to insert a node at the beginning of``       ``the Doubly Linked List */``    ``void` `push(``int` `new_data)``    ``{``        ``/* allocate node */``        ``LNode new_node = ``new` `LNode(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(LNode node)``    ``{``        ``while` `(node != ``null``)``        ``{``            ``System.out.print(node.data + ``" "``);``            ``node = node.next;``        ``}``    ``}` `    ``/* A utility function to print preorder traversal of BST */``    ``void` `preOrder(TNode node)``    ``{``        ``if` `(node == ``null``)``            ``return``;``        ``System.out.print(node.data + ``" "``);``        ``preOrder(node.left);``        ``preOrder(node.right);``    ``}` `    ``/* Driver 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(head);` `        ``/* Convert List to BST */``        ``TNode 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)`

## Python3

 `# Python3 implementation of above approach` `# Link list node``class` `LNode :``    ``def` `__init__(``self``):``        ``self``.data ``=` `None``        ``self``.``next` `=` `None` `# A Binary Tree node``class` `TNode :``    ``def` `__init__(``self``):``        ``self``.data ``=` `None``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `head ``=` `None` `# This function counts the number of``# nodes in Linked List and then calls``# sortedListToBSTRecur() to construct BST``def` `sortedListToBST():``    ``global` `head``    ` `    ``# Count the number of nodes in Linked List``    ``n ``=` `countLNodes(head)` `    ``# Construct BST``    ``return` `sortedListToBSTRecur(n)` `# The main function that constructs``# balanced BST and returns root of it.``# head -. Pointer to pointer to``# head node of linked list n -. No.``# of nodes in Linked List``def` `sortedListToBSTRecur( n) :``    ``global` `head``    ` `    ``# Base Case``    ``if` `(n <``=` `0``) :``        ``return` `None` `    ``# Recursively construct the left subtree``    ``left ``=` `sortedListToBSTRecur( ``int``(n``/``2``))` `    ``# Allocate memory for root, and``    ``# link the above constructed left``    ``# subtree with root``    ``root ``=` `newNode((head).data)``    ``root.left ``=` `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) which is n-n/2-1``    ``root.right ``=` `sortedListToBSTRecur( n ``-` `int``(n``/``2``) ``-` `1``)` `    ``return` `root` `# UTILITY FUNCTIONS` `# A utility function that returns``# count of nodes in a given Linked List``def` `countLNodes(head) :` `    ``count ``=` `0``    ``temp ``=` `head``    ``while``(temp !``=` `None``):``    ` `        ``temp ``=` `temp.``next``        ``count ``=` `count ``+` `1``    ` `    ``return` `count` `# Function to insert a node``#at the beginning of the linked list``def` `push(head, new_data) :` `    ``# allocate node``    ``new_node ``=` `LNode()``    ` `    ``# put in the data``    ``new_node.data ``=` `new_data` `    ``# link the old list off the new node``    ``new_node.``next` `=` `(head)` `    ``# move the head to point to the new node``    ``(head) ``=` `new_node``    ``return` `head`  `# Function to print nodes in a given linked list``def` `printList(node):` `    ``while``(node !``=` `None``):``    ` `        ``print``( node.data ,end``=` `" "``)``        ``node ``=` `node.``next``    ` `# Helper function that allocates a new node with the``# given data and None left and right pointers.``def` `newNode(data) :` `    ``node ``=` `TNode()``    ``node.data ``=` `data``    ``node.left ``=` `None``    ``node.right ``=` `None` `    ``return` `node` `# A utility function to``# print preorder traversal of BST``def` `preOrder( node) :` `    ``if` `(node ``=``=` `None``) :``        ``return``    ``print``(node.data, end ``=` `" "` `)``    ``preOrder(node.left)``    ``preOrder(node.right)` `# Driver code` `# Start with the empty list``head ``=` `None` `# Let us create a sorted linked list to test the functions``# Created linked list will be 1.2.3.4.5.6.7``head ``=` `push(head, ``7``)``head ``=` `push(head, ``6``)``head ``=` `push(head, ``5``)``head ``=` `push(head, ``4``)``head ``=` `push(head, ``3``)``head ``=` `push(head, ``2``)``head ``=` `push(head, ``1``)` `print``(``"Given Linked List "` `)``printList(head)` `# Convert List to BST``root ``=` `sortedListToBST()``print``(``"\nPreOrder Traversal of constructed BST "``)``preOrder(root)` `# This code is contributed by Arnab Kundu`

## C#

 `// C# implementation of above approach``using` `System;``    ` `public` `class` `LinkedList``{` `    ``/* head node of link list */``    ``static` `LNode head;``    ` `    ``/* Link list Node */``    ``class` `LNode``    ``{``        ``public` `int` `data;``        ``public` `LNode next, prev;` `        ``public` `LNode(``int` `d)``        ``{``            ``data = d;``            ``next = prev = ``null``;``        ``}``    ``}``    ` `    ``/* A Binary Tree Node */``    ``class` `TNode``    ``{``        ``public` `int` `data;``        ``public` `TNode left, right;` `        ``public` `TNode(``int` `d)``        ``{``            ``data = d;``            ``left = right = ``null``;``        ``}``    ``}` `    ``/* This function counts the number``    ``of nodes in Linked List and then calls``      ``sortedListToBSTRecur() to construct BST */``    ``TNode 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 */``    ``TNode sortedListToBSTRecur(``int` `n)``    ``{``        ``/* Base Case */``        ``if` `(n <= 0)``            ``return` `null``;` `        ``/* Recursively construct the left subtree */``        ``TNode left = sortedListToBSTRecur(n / 2);` `        ``/* head_ref now refers to middle node,``        ``make middle node as root of BST*/``        ``TNode root = ``new` `TNode(head.data);` `        ``// Set pointer to left subtree``        ``root.left = 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.right = sortedListToBSTRecur(n - n / 2 - 1);` `        ``return` `root;``    ``}` `    ``/* UTILITY FUNCTIONS */``    ``/* A utility function that returns count ``    ``of nodes in a given Linked List */``    ``int` `countNodes(LNode head)``    ``{``        ``int` `count = 0;``        ``LNode temp = head;``        ``while` `(temp != ``null``)``        ``{``            ``temp = temp.next;``            ``count++;``        ``}``        ``return` `count;``    ``}` `    ``/* Function to insert a node at the beginning of``    ``the Doubly Linked List */``    ``void` `push(``int` `new_data)``    ``{``        ``/* allocate node */``        ``LNode new_node = ``new` `LNode(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(LNode node)``    ``{``        ``while` `(node != ``null``)``        ``{``            ``Console.Write(node.data + ``" "``);``            ``node = node.next;``        ``}``    ``}` `    ``/* A utility function to print``    ``preorder traversal of BST */``    ``void` `preOrder(TNode node)``    ``{``        ``if` `(node == ``null``)``            ``return``;``        ``Console.Write(node.data + ``" "``);``        ``preOrder(node.left);``        ``preOrder(node.right);``    ``}` `    ``/* Driver code */``    ``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);` `        ``Console.WriteLine(``"Given Linked List "``);``        ``llist.printList(head);` `        ``/* Convert List to BST */``        ``TNode root = llist.sortedListToBST();``        ``Console.WriteLine(``""``);``        ``Console.WriteLine(``"Pre-Order Traversal of constructed BST "``);``        ``llist.preOrder(root);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:
```Given Linked List 1 2 3 4 5 6 7
PreOrder Traversal of constructed BST 4 2 1 3 6 5 7```

Time Complexity: O(n)