# Sorted Array to Balanced BST

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
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Algorithm
In the previous post, we discussed 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++ program to print BST in given range ` `#include ` `using` `namespace` `std; ` ` `  `/* A Binary Tree node */` `class` `TNode  ` `{  ` `    ``public``: ` `    ``int` `data;  ` `    ``TNode* left;  ` `    ``TNode* right;  ` `};  ` ` `  `TNode* newNode(``int` `data);  ` ` `  `/* A function that constructs Balanced ` `Binary 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 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()  ` `{  ` `    ``int` `arr[] = {1, 2, 3, 4, 5, 6, 7};  ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);  ` ` `  `    ``/* 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 `

## C

 `#include ` `#include ` ` `  `/* 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); ` ` `  `    ``/* Convert List to BST */` `    ``struct` `TNode *root = sortedArrayToBST(arr, 0, n-1); ` `    ``printf``(``"n PreOrder Traversal of constructed BST "``); ` `    ``preOrder(root); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to print BST in given range ` ` `  `// A binary tree node ` `class` `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

 `# Python code to convert a sorted array ` `# to a balanced Binary Search Tree ` ` `  `# binary tree node ` `class` `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 BST ` `def` `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.right ``=` `sortedArrayToBST(arr[mid``+``1``:]) ` `    ``return` `root ` ` `  `# A utility function to print the preorder  ` `# traversal of the BST ` `def` `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  `

## C#

 `using` `System; ` ` `  `// C# program to print BST in given range  ` ` `  `// A binary tree node  ` `public` `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 `

Output:

```Preorder traversal of constructed BST
4 2 1 3 6 5 7
```

Time Complexity: O(n)
Following is the recurrance 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.