Implementing tree with Ruby

Question

As regards in Ruby we don't have pointer like c++ , How can we implement tree?

Answer 1

You don't necessarily need pointers or references for building trees, do you?

Here is a basic example:

class Tree
  attr_accessor :children, :value

  def initialize(v)
    @value = v
    @children = []
  end
end

t = Tree.new(7)
t.children << Tree.new(3)
t.children << Tree.new(11)

t.value              # 7
t.children[0].value  # 3
t.children[1].value  # 11

Answer 2

One doesn't really need explicit pointers. Though people may expect this because they often learn about self-referential data structures in C and C++ and expect to see the equivalent of a point with an * sign preceding it. I believe the following snippet might be useful.

Ref: http://www.thelearningpoint.net/computer-science/basic-data-structures-in-ruby---binary-search-tre

# Example of Self-Referential Data Structures - A Binary Tree

class TreeNode
    attr_accessor :value, :left, :right

    # The Tree node contains a value, and a pointer to two children - left and right 
    # Values lesser than this node will be inserted on its left
    # Values greater than it will be inserted on its right
    def initialize val,left,right
        @value = val
        @left = left
        @right = right
    end
end

class BinarySearchTree

    # Initialize the Root Node
    def initialize val
        puts "Initializing with: " + val.to_s
        @root = TreeNode.new(val,nil,nil)   
    end

    # Pre-Order Traversal
    def preOrderTraversal(node= @root)
        return if (node == nil)
        preOrderTraversal(node.left)
        preOrderTraversal(node.right)
        puts node.value.to_s
    end

    # Post-Order Traversal
    def postOrderTraversal(node = @root)
        return if (node == nil)
        puts node.value.to_s
        postOrderTraversal(node.left)
        postOrderTraversal(node.right)
    end

    # In-Order Traversal : Displays the final output in sorted order
    # Display smaller children first (by going left)
    # Then display the value in the current node 
    # Then display the larger children on the right
    def inOrderTraversal(node = @root)
        return if (node == nil)
        inOrderTraversal(node.left)
        puts node.value.to_s
        inOrderTraversal(node.right)
    end


    # Inserting a value
    # When value > current node, go towards the right
    # when value < current node, go towards the left
    # when you hit a nil node, it means, the new node should be created there
    # Duplicate values are not inserted in the tree
    def insert(value)
        puts "Inserting :" + value.to_s
        current_node = @root
        while nil != current_node
            if (value < current_node.value) && (current_node.left == nil)
                current_node.left = TreeNode.new(value,nil,nil)
            elsif  (value > current_node.value) && (current_node.right == nil)
                current_node.right = TreeNode.new(value,nil,nil)
            elsif (value < current_node.value)
                current_node = current_node.left
            elsif (value > current_node.value)
                current_node = current_node.right
            else
                return
            end
        end
    end
end

bst = BinarySearchTree.new(10)
bst.insert(11)
bst.insert(9)
bst.insert(5)
bst.insert(7)
bst.insert(18)
bst.insert(17)
# Demonstrating Different Kinds of Traversals
puts "In-Order Traversal:"
bst.inOrderTraversal
puts "Pre-Order Traversal:"
bst.preOrderTraversal
puts "Post-Order Traversal:"
bst.postOrderTraversal

=begin

Output :
Initializing with: 10
Inserting :11
Inserting :9
Inserting :5
Inserting :7
Inserting :18
Inserting :17
In-Order Traversal:
5
7
9
10
11
17
18
Pre-Order Traversal:
7
5
9
17
18
11
10
Post-Order Traversal:
10
9
5
7
11
18
17

=end