Sorted Linked List to Balanced BST

24 Jun, 2022
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 


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

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

Input:  Linked List 1->2->3->4->5->6
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.  
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
Java
Python3
C#
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>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<<node->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
#include<stdio.h>#include<stdlib.h>  /* 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;}
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 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# implementation of above approachusing 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
<script>   // JavaScript implementation of above approach/* head node of link list */var head = null;  /* Link list Node */class LNode {    constructor(d)    {        this.data = d;        this.next = null;        this.prev = null;    }}  /* A Binary Tree Node */class TNode {    constructor(d)    {        this.data = d;        this.left = null;        this.right = null;    }}  /* This function counts the numberof nodes in Linked List and then calls  sortedListToBSTRecur() to construct BST */function sortedListToBST() {    /*Count the number of nodes in Linked List */    var 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 */function sortedListToBSTRecur(n) {    /* Base Case */    if (n <= 0)         return null;    /* Recursively construct the left subtree */    var left = sortedListToBSTRecur(parseInt(n / 2));    /* head_ref now refers to middle node,     make middle node as root of BST*/    var 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 - parseInt(n / 2) - 1);    return root;}/* UTILITY FUNCTIONS *//* A utility function that returns count  of nodes in a given Linked List */function countNodes(head) {    var count = 0;    var temp = head;    while (temp != null)    {        temp = temp.next;        count++;    }    return count;}  /* Function to insert a node at the beginning of the Doubly Linked List */function push( new_data) {    /* allocate node */    var 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 */function printList( node) {    while (node != null)     {        document.write(node.data + " ");        node = node.next;    }}  /* A utility function to print preorder traversal of BST */function preOrder(node) {    if (node == null)        return;    document.write(node.data + " ");    preOrder(node.left);    preOrder(node.right);}  /* Driver code */  /* Let us create a sorted linked list to test the functionsCreated linked list will be 7->6->5->4->3->2->1 */push(7);push(6);push(5);push(4);push(3);push(2);push(1);document.write("Given Linked List ");printList(head);/* Convert List to BST */var root = sortedListToBST();document.write("<br>");document.write("Pre-Order Traversal of constructed BST ");preOrder(root);  </script>
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)
Auxiliary Space: O(n) for call stack since using recursion
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