Re^2: Convert array to tree OR why variable changes arbitrarily

Thank you, this is elegant solution (a little bit complicated for late night but working :)). How did you mean keeping counter for $index? And also, I am aware that this will work only if @arr is "sorted", I will try to improve algorithm and if I lost then I will ask again :) Thank you.

Comment on Re^2: Convert array to tree OR why variable changes arbitrarily

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Re^3: Convert array to tree OR why variable changes arbitrarily by hdb (Monsignor) on May 31, 2013 at 07:11 UTC
Basically, you would be counting the `$treeElement`s while iterating over them, so you have the correct `$index` always at hand. Looks a bit redundant, but I prefer it over `grep`. `sub buildTree{ my ($tree, $parNg, $subNg) = @_; my $index = 0; for my $treeElement (@{$tree}){ if (ref $treeElement eq "ARRAY"){ return 1 if &buildTree($treeElement, $parNg, $subNg); } else { if ($treeElement eq $parNg){ splice @{$tree}, $index + 1, 0, $subNg; return 1; } } $index++; } return 0; }` [download]	[reply] [d/l] [select]

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Re^3: Convert array to tree OR why variable changes arbitrarily
by hdb (Monsignor) on May 31, 2013 at 07:11 UTC

Basically, you would be counting the $treeElements while iterating over them, so you have the correct $index always at hand. Looks a bit redundant, but I prefer it over grep.

sub buildTree{
  my ($tree, $parNg, $subNg) = @_;
  my $index = 0;
  for my $treeElement (@{$tree}){
    if (ref $treeElement eq "ARRAY"){
      return 1 if &buildTree($treeElement, $parNg, $subNg);
    } else {
      if ($treeElement eq $parNg){
    splice @{$tree}, $index + 1, 0, $subNg;
    return 1;
      }
    }
    $index++;
  }
  return 0;
}
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