ovedpo15 has asked for the wisdom of the Perl Monks concerning the following question:

Hi Monks!
I have the following sturcture:
{
  "tool": "A",
  "version": "1.3",
  "dependencies": [
    {
      "tool": "B",
      "version": "8.23",
      "dependencies": []
    },
    {
      "tool": "C",
      "version": "2.1",
      "dependencies": [
        {
          "tool": "D",
          "version": "1.1",
          "dependencies": []
        }
      ]
    }
  ]
}

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You can think of this data structure as a tree: 
  A
 / \
 B C
   |
   D

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I want to do a postorder traversal on this tree and to be able to print the tool/version. So I get: 
B/8.23
D/1.1
C/2.1
A/1.3

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The problem is that I'm not sure how to to do it. I need to find each leaf and then remove it (???) in order to mark that "I was here". I'm sure that there is a better way to handle it and this is why I'm asking here. Is it possible to suggest a better way?

Replies are listed 'Best First'.
Re: Postorder on hash structure
by Corion (Patriarch) on Aug 08, 2021 at 16:16 UTC

    Recursion is a good approach here.

    Consider the following example code that prints an array. You basically have two cases:

    1. The item passed does not have dependencies - print the item
    2. The item passed has references - for each reference, handle it by recursing, then print the item itself

    In rough code, this would look like the following:

    sub print_item_dependencies( $item ) {
    for my $dep (@{$item->{dependencies}}) {
        print_item_dependencies( $dep );
    }
    say join "/", $item->{tool}, $item->{version}
}

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Re: Postorder on hash structure
by tybalt89 (Monsignor) on Aug 08, 2021 at 17:07 UTC 
    #!/usr/bin/perl

use strict; # https://perlmonks.org/?node_id=11135697
use warnings;

my $structure = 
{
  "tool" => "A",
  "version" => "1.3",
  "dependencies" => [
    {
      "tool" => "B",
      "version" => "8.23",
      "dependencies" => []
    },
    {
      "tool" => "C",
      "version" => "2.1",
      "dependencies" => [
        {
          "tool" => "D",
          "version" => "1.1",
          "dependencies" => []
        }
      ]
    }
  ]
};

printsubstructure($structure);

sub printsubstructure
  {
  my $hashref = shift;
  printsubstructure($_) for @{ $hashref->{dependencies} };
  print "$hashref->{tool}/$hashref->{version}\n";
  }

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[d/l]
Re: Postorder on hash structure
by karlgoethebier (Abbot) on Aug 08, 2021 at 15:43 UTC
[reply]

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