Convert Sorted Array to Balanced Binary Search Tree (BST)

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

Solution:
If you would have to choose an array element to be the root of a balanced BST, which element would you pick? The root of a balanced BST should be the middle element from the sorted array.

You would pick the middle element from the sorted array in each iteration. You then create a node in the tree initialized with this element. After the element is chosen, what is left? Could you identify the sub-problems within the problem?

There are two arrays left — The one on its left and the one on its right. These two arrays are the sub-problems of the original problem, since both of them are sorted. Furthermore, they are subtrees of the current node’s left and right child.

The code below creates a balanced BST from the sorted array in O(N) time (N is the number of elements in the array). Compare how similar the code is to a binary search algorithm. Both are using the divide and conquer methodology.


BinaryTree* sortedArrayToBST(int arr[], int start, int end) {
if (start > end) return NULL;
// same as (start+end)/2, avoids overflow.
int mid = start + (end - start) / 2;
BinaryTree *node = new BinaryTree(arr[mid]);
node->left = sortedArrayToBST(arr, start, mid-1);
node->right = sortedArrayToBST(arr, mid+1, end);
return node;
}

BinaryTree* sortedArrayToBST(int arr[], int n) {
return sortedArrayToBST(arr, 0, n-1);
}