From: Adam Shelly <adam.shelly@...>
Date: 2008-06-13T05:45:01+09:00
Subject: Re: Trie data structure

On 6/12/08, Justin To <tekmc@hotmail.com> wrote:

There are a couple of issues with this code, here are some hints.

>  # Return true if value exists or nil if value D.N.E.
>  def child_value?(value)
>    if @children.empty?              # If @children is empty, then there are no children
>      return 'empty'

'empty' is not nil - it will act like a true, which is the opposite of
what the comments say this function should do.

>    else
>      # If one of the children contains the value then return true
>      @children.each do |child|
>        if(child.value==value)
>          return true
>        else
>          return nil
>        end
>      end

this code is only testing the value of the first child: it has a
return statement in both branches of the 'if', so it will exit after
the first test.

>    end
>  # End of def child_value?(value)
>  end
>
>  # Return the child that added the value if the value D.N.E. else
> return nil, the value already exists
>  # DO NOT ALTER OR CALL OUTSIDE OF CLASS--belongs to def add_number(value)
you can use the 'private' keyword here, which enforces 'do not call
outside of class'.

>  def add_digit(digit)
>    if(!child_value?(digit))
>      child = self<<(digit)
>      puts "#{digit} added"
>      return child
>    else
>      puts "#{digit}: already exists in the child."
>      return nil
>    end
>  end
>
>  def add_number(number)
>    current_node = self
>    number.to_s.each_byte do |byte|
>      puts "add_number #{byte.chr.to_i}"
>      puts "Nodeclass: " + current_node.class.to_s
>      current_node = current_node.add_digit(byte.chr.to_i)
>    end
>
as your comments say, If the digit already exists in the children,
add_digit will return nil.  In that case, you will set current_node to
nil here.  You probably want to set it to something else.

>  end
>
> # End of class Trie
> end
>

hope this helps,
-Adam