# Reverse nodes in a binary tree

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### Juan

Jun 3, 2017, 4:45:04 AM6/3/17
to SWI-Prolog
Hi !

I want to know how to reverse the nodes in a binary tree like this :
`[[[[[],1,[]],5,[]],7,[[],3,[[],4,[]]]],6,[[[],10,[]],8,[[[],9,[]],11,[[],2,[]]]]]`

``I` can only use this ABN representation as lists and the following constructors and selectors ofthe abstract data type.`empty([]).root([_,N,_], N).hi([HI,_,_],HI). %left sonhd([_,_,HD],HD). %right sonmaketree(R,HI,HD,[HI,R,HD}).`I want to know how could I make a swap predicate like this `swap(A4,B).` that working like this:`?- swap(A4,B). B=[[[[[],2,[]],11,[[],9,[]]],8,[[],10,[]]],6,[[[[],4,[]],3,[]],7,[[],5,[[],1,[]]]]]Any clues ?Thanks``

### Kuniaki Mukai

Jun 3, 2017, 5:38:57 AM6/3/17
to Juan, SWI-Prolog
Your problem sounds to me like this:

given binary tree
=> flatten the right most branch
=> reverse the flatten list
=> make it a binary tree

For example.
t(a,t(b, t(c)) =>
=> [a, b, c]
=> [c, b, a]
=> t(c, t(b, t(a))

Am I missing something ? Anyway, I hope this help.

Kuniaki Mukai

> On Jun 3, 2017, at 17:45, Juan <juaan...@gmail.com> wrote:
>
> Hi !
>
> I want to know how to reverse the nodes in a binary tree like this :
>
> [[[[[],1,[]],5,[]],
> 7,
> [[],3,[[],4,[]]]],
> 6,
> [[[],10,[]],
> 8,
> [[[],9,[]],11,[[],2,[]]]]]
>
>
> I can only use this ABN representation as lists and the following constructors and selectors of
> the abstract data type.
>
> empty([]).
> root([_,N,_], N).
> hi([HI,_,_],HI). %left son
> hd([_,_,HD],HD). %right son
> maketree(R,HI,HD,[HI,R,HD}).
>
>
> I want to know how could I make a swap predicate like this
> swap(A4,B).
>
> that working like this:
> ?- swap(A4,B).
> B=[[[[[],2,[]],11,[[],9,[]]],8,[[],10,[]]],6,[[[[],4,[]],3,[]],7,[[],5,[[],1,[]]]]]
>
> Any clues ?
>
> Thanks
>
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### Boris Vassilev

Jun 3, 2017, 7:51:49 AM6/3/17
to Kuniaki Mukai, Juan, SWI-Prolog
Hello Juan, Kuniaki,

I must be misunderstanding the question, because it seems this predicate does what you show, but not what you ask:

==
tree_mirror([], []). % the empty tree
tree_mirror([L, X, R], [Rm, X, Lm]) :- % swap the mirrored children
tree_mirror(L, Lm), % mirror the left...
tree_mirror(R, Rm). % ... and the right child

==

==
?- tree_mirror(

[[[[[],1,[]],5,[]],
7,
[[],3,[[],4,[]]]],
6,
[[[],10,[]],
8,
[[[],9,[]],11,[[],2,[]]]]],
R).
R = [[[[[], 2, []], 11, [[], 9, []]], 8, [[], 10, []]], 6, [[[[], 4, []], 3, []], 7, [[], 5, [[], 1, []]]]].

==

So at least I get what you seem to be after. I don't know if this is "reversing", I'd call it "mirroring". For reversing you need an order, so I guess in-order traversal? See Kuniaki's suggeston for a more general solution, however: if you are making a balanced binary tree from the reversed flattened list, as he suggested, you will **not** just mirror the tree in the general case. Rather, you will get a balanced tree that will have the nodes in reverse, if you do in-order traversal. I think doing this might be more useful than just "mirroring".

I also don't know how to use the "constructors and selectors of the abstract data type" without making the definition look clumsy :-(

Hope that helps more than it confuses,
Boris

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