You 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 also learned about tree search algorithms like breadth-first-search and depth-first-search. You learned that BFS is best used to find the optimum solution but can take a very long time (impractically long for broad and deep data sets) while DFS is often much faster but will give you the FIRST solution, not necessarily the best. Here you’ll get a chance to implement both.
You’ll build a simple binary tree data structure from some arbitrary input and also the “crawler” function that will locate data inside of it.
Node
. It should have a value
that it stores and also links to its parent and children (if they exist). Build getters and setters for it (e.g. parent node, child node(s)).build_tree
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 binary tree full of Node
objects appropriately placed. Start by assuming the array you get is sorted.build_tree
to handle data that isn’t presorted and cannot be easily sorted prior to building the tree. You’ll need to figure out how to add a node for each of the possible cases (e.g. if it’s a leaf versus in the middle somewhere).build_tree
so you can test it out.breadth_first_search
which takes a target value and returns the node at which it is located using the breadth first search technique. 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 search and to add new ones to the list (as you saw in the video). If the target node value is not located, return nil
.depth_first_search
which returns the node at which the target value is located using the depth first search technique. Use an array acting as a stack to do this.dfs_rec
which runs a depth first search as before but this time, instead of using a stack, make this method recursive.1. You can think of the `dfs_rec` method as a little robot that crawls down the tree, checking if a node is the correct node and spawning other little robots to keep searching the tree. No robot is allowed to turn on, though, until all the robots to its left have finished their task.
2. The method will need to take in both the target value and the current node to compare against.
Submit a link below to this file on the ruby course github repo with your files in it by using a pull request. See the section on Contributing for how.
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:
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]]
> knight_moves([3,3],[4,3])
=> You made it in 3 moves! Heres your path:
[3,3]
[4,5]
[2,4]
[4,3]
Send us your solution so we can show others! Submit a link to the Github repo with your files in it here using any of the methods listed on the contributing page. Please include your partner’s github handle somewhere in the description if they would like attribution.
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