Idea of converting array to BST

Let's convert a sorted array to BST

Steps:
Find the middle element of the array: This will be the root of the BST.

Recursively construct the left subtree: Use the left half of the array (elements before the middle element).

Recursively construct the right subtree: Use the right half of the array (elements after the middle element).

Return the root: The root will have its left and right children set to the roots of the left and right subtrees, respectively.

#include <bits/stdc++.h>
using namespace std;
class Node
{
public:
    int val;
    Node *right;
    Node *left;
    Node(int value)
    {
        this->val = value;
        this->right = NULL;
        this->left = NULL;
    }
};

void level_order(Node *root)
{
    if (root == NULL)
    {
        cout << "Tree nai" << endl;
        return;
    }
    queue<Node *> q;
    q.push(root);
    while (!q.empty())
    {
        // 1. ber kore ana
        Node *f = q.front();
        q.pop();

        // 2. jabotiyo ja kaj ache
        cout << f->val << " ";

        // 3. tar children gulo ke rakha
        if (f->left)
            q.push(f->left);
        if (f->right)
            q.push(f->right);
    }
}
Node *convert(int a[], int n, int l, int r)
{
    if (l > r)
        return NULL; // base case 

    int mid = (l + r) / 2;
    Node *root = new Node(a[mid]);
    Node *leftroot = convert(a, n, l, mid - 1);
    Node *rightroot = convert(a, n, mid + 1, r);

    root->left = leftroot;
    root->right = rightroot;

    return root;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n;
    cin >> n;
    int a[n];
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    Node *root = convert(a, n, 0, n - 1); 
    level_order(root); // Printing 
}