Because you can assign only one scalar value to a hash slot. So you can say $hash{tree} = 4; or you can say $hash{tree} = {apple => 6}; ¹, but not both at the same time.

¹: This which would assign a scalar to $hash{tree}, that scalar being a reference to the anonymous hash {apple=>6}. In a way this boils down to the same as $hash->{tree}->{apple} $hash{tree}->{apple} (Update: typo kindly pointed out by AnomalousMonk), see for yourself:

use strict;
use warnings;
use Data::Dump 'pp';

my %hash;

$hash{tree}->{apple} = 6;
pp \%hash;

$hash{tree} = {apple => 6};
pp \%hash;

__END__
{ tree => { apple => 6 } }
{ tree => { apple => 6 } }
[download]

If you really need a tree structure, perhaps a hash of arrays would be better suited. It'd get something like this:

use strict;
use warnings;
use Data::Dump 'pp';

my %hash;

$hash{tree} = [4];
push @{$hash{tree}}, {apple => [6]};

pp \%hash;

my $apple_tree = $hash{tree}->[1]->{apple};
push @$apple_tree, {red => 9, fuju => 4};

pp \%hash;

__END__
{ tree => [4, { apple => [6] }] }
{ tree => [4, { apple => [6, { fuju => 4, red => 9 }] }] }
[download]

Or, if you really want your own code to walk over this tree:

for my $key (keys %hash) {
    my ($root_value, $subtree) = @{$hash{$key}};
    print "$key  $root_value\n";

    for my $subtree_key (keys %$subtree) {
        my ($subtree_root_value, $subtree_subtree) = @{$subtree->{$sub
+tree_key}};
        print "$key  $subtree_key  $subtree_root_value\n";

        for my $bottom_level_key (keys %$subtree_subtree) {
            print "$key  $subtree_key  $bottom_level_key  ", $subtree_
+subtree->{$bottom_level_key}, "\n";
        }
    }
}

__END__
tree  4
tree  apple  6
tree  apple  fuju  4
tree  apple  red  9
[download]

But as you can see that becomes pretty messy rather quickly, codewise. Perhaps you should reconsider the data structure you want to use?

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