Sorted Array to Balanced BST

17 Jun, 2022
Given a sorted array. Write a function that creates a Balanced Binary Search Tree using array elements.Examples:  
Input:  Array {1, 2, 3}
Output: A Balanced BST
     2
   /  \
  1    3 

Input: Array {1, 2, 3, 4}
Output: A Balanced BST
      3
    /  \
   2    4
 /
1
Algorithm: In the previous post, we discussed the construction of BST from sorted Linked List. Constructing from sorted array in O(n) time is simpler as we can get the middle element in O(1) time. 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 array 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.
Following is the implementation of the above algorithm. The main code which creates Balanced BST is highlighted.
C++
C
Java
Python
C#
Javascript
// C++ program to print BST in given range#include<bits/stdc++.h>using namespace std; /* A Binary Tree node */class TNode{    public:    int data;    TNode* left;    TNode* right;}; TNode* newNode(int data); /* A function that constructs BalancedBinary Search Tree from a sorted array */TNode* sortedArrayToBST(int arr[],                        int start, int end){    /* Base Case */    if (start > end)    return NULL;     /* Get the middle element and make it root */    int mid = (start + end)/2;    TNode *root = newNode(arr[mid]);     /* Recursively construct the left subtree    and make it left child of root */    root->left = sortedArrayToBST(arr, start,                                    mid - 1);     /* Recursively construct the right subtree    and make it right child of root */    root->right = sortedArrayToBST(arr, mid + 1, end);     return root;} /* Helper function that allocates a new nodewith the given data and NULL left and rightpointers. */TNode* newNode(int data){    TNode* node = new TNode();    node->data = data;    node->left = NULL;    node->right = NULL;     return node;} /* A utility function to printpreorder traversal of BST */void preOrder(TNode* node){    if (node == NULL)        return;    cout << node->data << " ";    preOrder(node->left);    preOrder(node->right);} // Driver Codeint main(){    int arr[] = {1, 2, 3, 4, 5, 6, 7};    int n = sizeof(arr) / sizeof(arr[0]);     /* Convert List to BST */    TNode *root = sortedArrayToBST(arr, 0, n-1);    cout << "PreOrder Traversal of constructed BST \n";    preOrder(root);     return 0;} // This code is contributed by rathbhupendra
#include<stdio.h>#include<stdlib.h> /* A Binary Tree node */struct TNode{    int data;    struct TNode* left;    struct TNode* right;}; struct TNode* newNode(int data); /* A function that constructs Balanced Binary Search Tree from a sorted array */struct TNode* sortedArrayToBST(int arr[], int start, int end){    /* Base Case */    if (start > end)      return NULL;     /* Get the middle element and make it root */    int mid = (start + end)/2;    struct TNode *root = newNode(arr[mid]);     /* Recursively construct the left subtree and make it       left child of root */    root->left =  sortedArrayToBST(arr, start, mid-1);     /* Recursively construct the right subtree and make it       right child of root */    root->right = sortedArrayToBST(arr, mid+1, end);     return root;} /* 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(){    int arr[] = {1, 2, 3, 4, 5, 6, 7};    int n = sizeof(arr)/sizeof(arr[0]);     /* Convert List to BST */    struct TNode *root = sortedArrayToBST(arr, 0, n-1);    printf("n PreOrder Traversal of constructed BST ");    preOrder(root);     return 0;}
// Java program to print BST in given range // A binary tree nodeclass Node {         int data;    Node left, right;         Node(int d) {        data = d;        left = right = null;    }} class BinaryTree {         static Node root;     /* A function that constructs Balanced Binary Search Tree     from a sorted array */    Node sortedArrayToBST(int arr[], int start, int end) {         /* Base Case */        if (start > end) {            return null;        }         /* Get the middle element and make it root */        int mid = (start + end) / 2;        Node node = new Node(arr[mid]);         /* Recursively construct the left subtree and make it         left child of root */        node.left = sortedArrayToBST(arr, start, mid - 1);         /* Recursively construct the right subtree and make it         right child of root */        node.right = sortedArrayToBST(arr, mid + 1, end);                 return node;    }     /* A utility function to print preorder traversal of BST */    void preOrder(Node node) {        if (node == null) {            return;        }        System.out.print(node.data + " ");        preOrder(node.left);        preOrder(node.right);    }         public static void main(String[] args) {        BinaryTree tree = new BinaryTree();        int arr[] = new int[]{1, 2, 3, 4, 5, 6, 7};        int n = arr.length;        root = tree.sortedArrayToBST(arr, 0, n - 1);        System.out.println("Preorder traversal of constructed BST");        tree.preOrder(root);    }} // This code has been contributed by Mayank Jaiswal
# Python code to convert a sorted array# to a balanced Binary Search Tree # binary tree nodeclass Node:    def __init__(self, d):        self.data = d        self.left = None        self.right = None # function to convert sorted array to a# balanced BST# input : sorted array of integers# output: root node of balanced BSTdef sortedArrayToBST(arr):         if not arr:        return None     # find middle    mid = (len(arr)) / 2         # make the middle element the root    root = Node(arr[mid])         # left subtree of root has all    # values <arr[mid]    root.left = sortedArrayToBST(arr[:mid])         # right subtree of root has all    # values >arr[mid]    root.right = sortedArrayToBST(arr[mid+1:])    return root # A utility function to print the preorder# traversal of the BSTdef preOrder(node):    if not node:        return         print node.data,    preOrder(node.left)    preOrder(node.right) # driver program to test above function"""Constructed balanced BST is    4/ \2 6/ \ / \1 3 5 7""" arr = [1, 2, 3, 4, 5, 6, 7]root = sortedArrayToBST(arr)print "PreOrder Traversal of constructed BST ",preOrder(root) # This code is contributed by Ishita Tripathi
using System; // C# program to print BST in given range // A binary tree nodepublic class Node{     public int data;    public Node left, right;     public Node(int d)    {        data = d;        left = right = null;    }} public class BinaryTree{     public static Node root;     /* A function that constructs Balanced Binary Search Tree      from a sorted array */    public virtual Node sortedArrayToBST(int[] arr, int start, int end)    {         /* Base Case */        if (start > end)        {            return null;        }         /* Get the middle element and make it root */        int mid = (start + end) / 2;        Node node = new Node(arr[mid]);         /* Recursively construct the left subtree and make it         left child of root */        node.left = sortedArrayToBST(arr, start, mid - 1);         /* Recursively construct the right subtree and make it         right child of root */        node.right = sortedArrayToBST(arr, mid + 1, end);         return node;    }     /* A utility function to print preorder traversal of BST */    public virtual void preOrder(Node node)    {        if (node == null)        {            return;        }        Console.Write(node.data + " ");        preOrder(node.left);        preOrder(node.right);    }     public static void Main(string[] args)    {        BinaryTree tree = new BinaryTree();        int[] arr = new int[]{1, 2, 3, 4, 5, 6, 7};        int n = arr.Length;        root = tree.sortedArrayToBST(arr, 0, n - 1);        Console.WriteLine("Preorder traversal of constructed BST");        tree.preOrder(root);    }}   // This code is contributed by Shrikant13
<script>   // JavaScript program to print BST in given range // A binary tree nodeclass Node{    constructor(d)    {        this.data = d;        this.left = null;        this.right = null;    }} var root = null; /* A function that constructs Balanced Binary Search Tree  from a sorted array */function sortedArrayToBST(arr, start, end){    /* Base Case */    if (start > end)    {        return null;    }    /* Get the middle element and make it root */    var mid = parseInt((start + end) / 2);    var node = new Node(arr[mid]);    /* Recursively construct the left subtree and make it     left child of root */    node.left = sortedArrayToBST(arr, start, mid - 1);    /* Recursively construct the right subtree and make it     right child of root */    node.right = sortedArrayToBST(arr, mid + 1, end);    return node;}/* 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);}  var arr = [1, 2, 3, 4, 5, 6, 7];var n = arr.length;root = sortedArrayToBST(arr, 0, n - 1);document.write("Preorder traversal of constructed BST<br>");preOrder(root); </script>
Output:
Preorder traversal of constructed BST
4 2 1 3 6 5 7 
Tree representation of above output:
     4  
 2      6
1  3  5   7
Time Complexity: O(n) 
Following is the recurrence relation for sortedArrayToBST(). 
  T(n) = 2T(n/2) + C
  T(n) -->  Time taken for an array of size n
   C   -->  Constant (Finding middle of array and linking root to left 
                      and right subtrees take constant time) 
The above recurrence can be solved using Master Theorem as it falls in case 1.  
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