Convert Sorted Array to Balanced BST in DSA Javascript - Build from Scratch

Convert Sorted Array to Balanced BST

📖 Scenario: You have a sorted list of numbers representing data that needs to be organized for fast searching. A balanced Binary Search Tree (BST) is a great way to do this because it keeps the data sorted and allows quick lookups.Imagine you are building a phone book app that needs to quickly find contacts by their phone numbers. You want to convert the sorted list of phone numbers into a balanced BST to make searching efficient.

🎯 Goal: Build a balanced Binary Search Tree (BST) from a given sorted array of numbers. The BST should be balanced, meaning the left and right subtrees of every node differ in height by no more than one.You will write code to create tree nodes, pick the middle element as the root, and recursively build left and right subtrees.

📋 What You'll Learn

Create a TreeNode class with val, left, and right properties

Use a recursive function sortedArrayToBST that takes a sorted array and returns the root of the balanced BST

Pick the middle element of the current array/subarray as the root node

Recursively build left subtree from the left half of the array

Recursively build right subtree from the right half of the array

Print the BST nodes in preorder traversal (root, left, right) to verify the structure

💡 Why This Matters

🌍 Real World

Balanced BSTs are used in databases, file systems, and search engines to organize data for fast searching and insertion.

💼 Career

Understanding how to convert sorted data into balanced trees is important for software engineers working on efficient data storage, retrieval, and algorithms.

1

Create the TreeNode class

Create a class called TreeNode with a constructor that takes a parameter val and sets this.val to it. Also initialize this.left and this.right to null.

DSA Javascript

# Your code here

Hint

Think of each tree node as a box holding a value and two empty slots for left and right child nodes.

2

Set up the sorted array

Create a constant array called nums with these exact values: [1, 2, 3, 4, 5, 6, 7].

DSA Javascript

class TreeNode {
    constructor(val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

// Create the sorted array nums
// Your code here

Hint

This array is already sorted, which is perfect for building a balanced BST.

3

Write the recursive function to build the BST

Write a function called sortedArrayToBST that takes a parameter arr. Inside, if arr.length is 0, return null. Otherwise, find the middle index as Math.floor(arr.length / 2). Create a TreeNode with the middle element. Recursively assign root.left by calling sortedArrayToBST on the left half of the array, and root.right by calling it on the right half. Return the root node.

DSA Javascript

class TreeNode {
    constructor(val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

const nums = [1, 2, 3, 4, 5, 6, 7];

// Write the sortedArrayToBST function
// Your code here

Hint

Use the middle element as the root to keep the tree balanced. Use array slicing to get left and right parts.

4

Print the BST nodes in preorder traversal

Write a function called preorderTraversal that takes a node. If node is null, return an empty array. Otherwise, return an array with node.val followed by the result of preorderTraversal(node.left) and then preorderTraversal(node.right). Then, create a variable root by calling sortedArrayToBST(nums). Finally, print the result of preorderTraversal(root) using console.log.

DSA Javascript

class TreeNode {
    constructor(val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

const nums = [1, 2, 3, 4, 5, 6, 7];

function sortedArrayToBST(arr) {
    if (arr.length === 0) return null;
    const mid = Math.floor(arr.length / 2);
    const root = new TreeNode(arr[mid]);
    root.left = sortedArrayToBST(arr.slice(0, mid));
    root.right = sortedArrayToBST(arr.slice(mid + 1));
    return root;
}

// Write preorderTraversal function and print the preorder traversal of the BST
// Your code here

Hint

Preorder traversal visits root first, then left subtree, then right subtree. Use recursion to collect values.