Re: iterative copying of binary expression trees

It is iterative and functional, I pass the stack as an argument.
push and pop are *not* destructive. See the last reply from pjb...

Thanks Pascal, I'll get an eye on your lap-level solution :-)

JG

In article
<a868c224-0358-42ee-a58f-0349999d702d@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
gugamilare <gugamilare@xxxxxxxxx> wrote:

On 23 maio, 12:01, Jean Guillaume Pyraksos <wis...@xxxxxxxxxxx> wrote:
None, I do *want* a tail-recursive, purely functional, first order solution.
I'll try myself, or prove me that this is impossible.
I have already such a tail-recursive, purely functional, first order solution
in O(n) but two-phase, in 2n iterative steps.
I'm looking for a one-phase solution. I don't see on the theoretical side why
I couldn't find one.

But are you sure that the first phase of your algorithm is iterative
and purely functional? You said you push the left son into a stack,
and this is not exactly functional (you need to use "pop", which is
destructive). You can do the exact same thing to create your algorithm.
.

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