Project: Data Structures and Algorithms

Project 1: Binary Search Trees

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 breadth-first and depth-first.

Now, let’s take a look at balanced binary search trees (BST). 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 1

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.

  1. Build a Node class. It is should have attributes for the data it stores as well as its left and right children. As a bonus, try including the Comparable module and make nodes compare using their data attribute.

  2. Build a Tree class which accepts an array when initialized. The Tree class should have a root attribute which uses the return value of #build_tree which you’ll write next.

  3. 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 of Node objects appropriately placed (don’t forget to sort and remove duplicates!). The #build_tree method should return the level-1 root node.

  4. 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.

  5. Write a #find method which accepts a value and returns the node with the given value.

  6. Write a #level_order method that returns an array of values. This method should traverse the tree in breadth-first level order. This method can be implemented using either iteration or recursion (try implementing both!). 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).

  7. Write #inorder, #preorder, and #postorder methods that returns an array of values. Each method should traverse the tree in their respective depth-first order.

  8. Write a #depth method which accepts a node and returns the depth(number of levels) beneath the node.

  9. 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.

  10. Write a #rebalance method which rebalances an unbalanced tree. Tip: You’ll want to create a level-order array of the tree before passing the array back into the #build_tree method.

  11. Write a simple driver script that does the following:

1. Create a binary search tree from an array of random numbers (` { rand(1..100) }`)
2. Confirm that the tree is balanced by calling `#balanced?`
3. Print out all elements in level, pre, post, and in order
4. try to unbalance the tree by adding several numbers > 100
5. Confirm that the tree is unbalanced by calling `#balanced?`
6. Balance the tree by calling `#rebalance`
7. Confirm that the tree is balanced by calling `#balanced?`
8. Print out all elements in level, pre, post, and in order

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 ? "└── " : "┌── "}#{}"
  pretty_print(node.left, "#{prefix}#{is_left ? " " : "│ "}", true) if node.left

Student Solutions

Submit a pull request with a link to your solution in this file when your project is complete. See the section on Contributing for details on how.

Show Student Solutions

Project 2: Knight’s Travails

Now you’re a pro with DFS and BFS. Let’s try using our search algorithms on a real problem.

For this project, you’ll need to use a data structure that’s similar (but not identical) to a binary tree. For a summary of a few different examples, reference this article.

A knight in chess can move to any square on the standard 8x8 chess board from any other square on the board, given enough turns (don’t believe it? See this animation). Its basic move is two steps forward and one step to the side. It can face any direction.

All the possible places you can end up after one move look like this:

Assignment 2

Your task is to build a function knight_moves that shows the simplest possible way to get from one square to another by outputting all squares the knight will stop on along the way.

You can think of the board as having 2-dimensional coordinates. Your function would therefore look like:

  • knight_moves([0,0],[1,2]) == [[0,0],[1,2]]
  • knight_moves([0,0],[3,3]) == [[0,0],[1,2],[3,3]]
  • knight_moves([3,3],[0,0]) == [[3,3],[1,2],[0,0]]
  1. Put together a script that creates a game board and a knight.
  2. Treat all possible moves the knight could make as children in a tree. Don’t allow any moves to go off the board.
  3. Decide which search algorithm is best to use for this case. Hint: one of them could be a potentially infinite series.
  4. Use the chosen search algorithm to find the shortest path between the starting square (or node) and the ending square. Output what that full path looks like, e.g.:
  > knight_moves([3,3],[4,3])
  => You made it in 3 moves!  Here's your path:

Student Solutions

Submit a pull request with a link to your solution in this file when your project is complete. See the section on Contributing for details on how.

Show Student Solutions
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