Convert Sorted Array to Binary Search Tree - Java Solution

1. Introduction

In this blog post, we will address a common problem in data structures: converting a sorted array into a height-balanced Binary Search Tree (BST). This problem is a good exercise in understanding how binary trees and binary search can be implemented in Java.

Problem

Given an array where elements are sorted in ascending order, convert it to a height-balanced BST, which is defined as a binary tree in which the depth of the two subtrees of every node never differs by more than 1.

Example:

Given the sorted array [1, 2, 3, 4, 5, 6]

One possible answer is:
      4
    /   \
   2     5
  / \     \
 1   3     6  

2. Solution Steps

1. The basic idea is to find the middle element of the array and make it the root of the BST.

2. The left half of the array will form the left subtree, and the right half will form the right subtree.

3. Recursively apply this process to the left and right halves of the array.

4. Continue this process until the entire array is converted into a height-balanced BST.

3. Code Program

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

public class Solution {

    // Main method for testing
    public static void main(String[] args) {
        int[] nums = {1, 2, 3, 4, 5, 6};
        TreeNode root = sortedArrayToBST(nums);
        printInOrder(root);
    }

    // Method to convert sorted array to BST
    public static TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null || nums.length == 0) return null; // Base case for empty array
        return constructBSTRecursive(nums, 0, nums.length - 1); // Construct BST recursively
    }

    // Helper method to construct BST
    private static TreeNode constructBSTRecursive(int[] nums, int left, int right) {
        if (left > right) return null; // Base case

        int mid = left + (right - left) / 2; // Middle element of the array
        TreeNode node = new TreeNode(nums[mid]); // Create a node with the middle element

        // Construct the left and right subtrees recursively
        node.left = constructBSTRecursive(nums, left, mid - 1);
        node.right = constructBSTRecursive(nums, mid + 1, right);

        return node; // Return the root node of the BST
    }

    // Method for in-order traversal of the BST
    public static void printInOrder(TreeNode node) {
        if (node == null) return; // Base case
        printInOrder(node.left); // Visit left subtree
        System.out.print(node.val + " "); // Print node value
        printInOrder(node.right); // Visit right subtree
    }
}

Output:

1 2 3 4 5 6

Explanation:

The sorted array [1, 2, 3, 4, 5, 6] is converted into a height-balanced BST. The middle element (4) becomes the root, [1, 2, 3] forms the left subtree, and [5, 6] forms the right subtree. The sortedArrayToBST function recursively constructs the BST and the printInOrder function prints the BST in an in-order traversal, which outputs the elements in the original sorted order.