[GeeksForGeeks] Sorted array to balanced BST

Given a sorted array. Write a program that creates a Balanced Binary Search Tree using array elements.

If there are n elements in array, then floor(n/2)'th element should be chosen as root and same should be followed recursively.

 

Solution.

1. get the middle element and create root node N.

2. recursively create a balanced BST from the left half of the array and set its root as N's left node.

3. recursively create a balanced BST from the right half of the array and set its root as N's right node.

 

T(n) = 2 * T(n/2) + O(1),  so the time complexity is O(n).

 

class TreeNode {
    TreeNode left;
    TreeNode right;
    int val;
    TreeNode(int val){
        this.left = null;
        this.right = null;
        this.val = val;
    }
}
public class Solution {
    public TreeNode sortedArrayToBalancedBST(int[] a) {
        if(a == null || a.length == 0){
            return null;
        }
        return toBSTRecursive(a, 0, a.length - 1);
    }
    private TreeNode toBSTRecursive(int[] a, int start, int end){
        if(start > end){
            return null;
        }
        int mid = start + (end - start) / 2;
        TreeNode node = new TreeNode(a[mid]);
        node.left = toBSTRecursive(a, start, mid - 1);
        node.right = toBSTRecursive(a, mid + 1, end);
        return node;
    }
}