import java.util.*;

/*
*	Given an array of distinct integers, use them to build a balanced binary search tree.
*/
class Solution {
	/*
	*	Node class for constructing a binary tree of integers.
	*/
	class Node {
		private int data;
		public Node left;
		public Node right;
		 	
		public Node(int data) {
			this.data = data;
			this.left = null;
			this.right = null;
		}
	}
	
	/*
	*	Build a binary search tree for a given range of indices. Called by {@link Solution#arrayToBalancedBST}.
	*	
	*	@param int[] numbers An array of distinct integers sorted in ascending order.
	*	@param int start The starting index for the smallest element in the BST.
	*	@param int end The ending index for the largest element in the BST.
	*
	*	@return Node The root of the BST constructed from the given range of indices.
	*/
	private Node build(int[] numbers, int start, int end) {
		System.err.println("Called build for range " + start + " to " + end);
		
		Node root = null;
		
		// If we can create at least one new node for the given range of indices
		if (start <= end) {
			// Get index of new parent/root node
			int mid = (start + end) / 2;
			root = new Node(numbers[mid]);
			System.err.println("Created Node: " + root.data);
			
			// This isn't super necessary but it saves 2 extra function 
			// calls for each leaf node that would return null
			if (start != end) { // Comment this out to see the unnecessary calls
				root.left = build(numbers, start, mid - 1);
				root.right = build(numbers, mid + 1, end);
			}
		}
		
		// Return the new root
		return root;
	}
	
	/*
	*	@return The root of a balanced binary search tree
	*/
	public Node arrayToBalancedBST(int[] numbers) {
		// Sort numbers in ascending order
		Arrays.sort(numbers);
		
		return build(numbers, 0, numbers.length - 1);
	}
	
	/*
	*	Prints an in-order tree traversal.
	*/
	public void inOrder(Node root) {
		// Traverse all the way left
		if (root.left != null) {
			inOrder(root.left);
		}
		
		// Print current node
		System.out.println("Traversed Node: " + root.data);
		
		// Continue traversal with right subtree
		if (root.right != null) {
			inOrder(root.right);
		}
	}
	
	/*
	*	Driver function
	*/
	public static void run() {
		Solution solution = new Solution();
		int[] numbers = {5, 4, 7, 3, 2, 1, 6};
		Node root = solution.arrayToBalancedBST(numbers);
		solution.inOrder(root);
	}
	
	public static void main(String[] args) {
		Solution.run();
	}
}