There are n! permutations that can be inserted into an initially empty binary tree, but only Catalan(n) different binary trees. A possible method (surely very ineffective, but I'm an old FORTRAN guy, meaning that I automatically hate classes and recursion) to weed out duplicates would be the following: If a left child is filled in and already a right child is present, duplicate (break loop), since in an "earlier" permutation, you already filled the left first, with the same result.
My question: The mirror of a binary tree has the same "structural" properties and I can exclude it too. But is a simple mirror argument enough to conclude I can loop only through the first half of the permutations, i.e. it can't be that I "forget" a tree because all the reversed permutations have been weeded out as above? Example: 123,132,213,231,312,321. 231 gets deleted, and that is "safe" as it is in the second half anyway.
To make it a bit more formal: Let $P$ be a permutation sequence, $P'$ its number conjugate (replace all $i$ by $n+1-i$, e.g. 213 by 231). Call $P$ "bad" if there exist $i<j<k$ such that $P(k)<P(i)<P(j)$. Call $P$ "small" if $P<P'$ lexicographically. Let $M=P(T)$ be the set of all $P$ that generate the binary tree $T$. Can it be that for some $T$ all small $P$ in $M$ are bad?
Or even shorter, courtesy of Gribouillis: Let $P$ be stack-sortable, see below. Can $P'$ (with $P'<P$, alphanumerically) be not stack-sortable?
EDIT: The formal question is correct, the informal not, as $T(3,1,4,2)=T(3,1,2,4)$ but $2$ and $4$ don't hang on the same node.
EDIT2: Argl. See my comment below.
