Introduction
You have learned about binary search trees – where you take a group of data items and turn them into a tree full of nodes where each left node is “lower” than each right node. The tree starts with the “root node” and any node with no children is called a “leaf node”. You have also learned about tree traversal algorithms like breadthfirst and depthfirst.
Now, let’s take a look at balanced binary search trees (BST). A BST allows fast operations for lookup, insertion, and deletion of data items. Read this article and watch this video to understand the basic algorithm used to build a balanced BST. Although these two resources do not use Ruby, you should understand it enough to develop your own pseudocode.
Assignment
You’ll build a balanced BST in this assignment. Do not use duplicate values because they make it more complicated and result in trees that are much harder to balance. Therefore, be sure to always remove duplicate values or check for an existing value before inserting.

Build a
Node
class. It should have an attribute for the data it stores as well as its left and right children. As a bonus, try including theComparable
module and compare nodes using their data attribute. 
Build a
Tree
class which accepts an array when initialized. TheTree
class should have aroot
attribute which uses the return value of#build_tree
which you’ll write next. 
Write a
#build_tree
method which takes an array of data (e.g. [1, 7, 4, 23, 8, 9, 4, 3, 5, 7, 9, 67, 6345, 324]) and turns it into a balanced binary tree full ofNode
objects appropriately placed (don’t forget to sort and remove duplicates!). The#build_tree
method should return the level0 root node.Tip: If you would like to visualize your binary search tree, here is a
#pretty_print
method that a student wrote and shared on Discord:def pretty_print(node = @root, prefix = '', is_left = true) pretty_print(node.right, "#{prefix}#{is_left ? '│ ' : ' '}", false) if node.right puts "#{prefix}#{is_left ? '└── ' : '┌── '}#{node.data}" pretty_print(node.left, "#{prefix}#{is_left ? ' ' : '│ '}", true) if node.left end

Write an
#insert
and#delete
method which accepts a value to insert/delete (you’ll have to deal with several cases for delete such as when a node has children or not). If you need additional resources, check out these two articles on inserting and deleting, or this video with several visual examples.You may be tempted to implement these methods using the original input array, but it’s important for the efficiency of these operations that you don’t do this. If we refer back to the Big O Cheatsheet, we’ll see that binary search trees can insert/delete in
O(log n)
time, which is a significant performance boost over arrays for the same operations. In order to get this added efficiency, your implementation of these methods should traverse the tree and manipulate the nodes and their connections. 
Write a
#find
method which accepts a value and returns the node with the given value. 
Write a
#level_order
method which accepts a block. This method should traverse the tree in breadthfirst level order and yield each node to the provided block. This method can be implemented using either iteration or recursion (try implementing both!). The method should return an array of values if no block is given. Tip: You will want to use an array acting as a queue to keep track of all the child nodes that you have yet to traverse and to add new ones to the list (as you saw in the video). 
Write
#inorder
,#preorder
, and#postorder
methods that accepts a block. Each method should traverse the tree in their respective depthfirst order and yield each node to the provided block. The methods should return an array of values if no block is given. 
Write a
#height
method which accepts a node and returns its height. Height is defined as the number of edges in longest path from a given node to a leaf node. 
Write a
#depth
method which accepts a node and returns its depth. Depth is defined as the number of edges in path from a given node to the tree’s root node. 
Write a
#balanced?
method which checks if the tree is balanced. A balanced tree is one where the difference between heights of left subtree and right subtree of every node is not more than 1. 
Write a
#rebalance
method which rebalances an unbalanced tree. Tip: You’ll want to use a traversal method to provide a new array to the#build_tree
method.
Tie it all together
Write a simple driver script that does the following:
 Create a binary search tree from an array of random numbers
(Array.new(15) { rand(1..100) })
 Confirm that the tree is balanced by calling
#balanced?
 Print out all elements in level, pre, post, and in order
 Unbalance the tree by adding several numbers > 100
 Confirm that the tree is unbalanced by calling
#balanced?
 Balance the tree by calling
#rebalance
 Confirm that the tree is balanced by calling
#balanced?
 Print out all elements in level, pre, post, and in order