Are reversed reverse preorder traversals equivalent to a postorder traversal?

I was viewing the solutions of other Leetcode users for the classic "post-order traversal of a binary tree" question, when to my surprise, I found a ton of users simply finding the reverse preorder traversal (because it is considerably easier to implement iteratively), and then reversing the output. To my further surprise, I could not find a single counterexample to these conjectures, which I will state clearly:

-
Conjecture 1: The post-order traversal of a tree $T$ is equivalent to the reversed reverse pre-order traversal of $T$.

-
Conjecture 2: The reverse post-order traversal of a tree $T$ is equivalent to the reversed pre-order traversal of $T$.

I thought about this plenty, and came up with an extremely informal justification for this: LRN = reverse(NRL) and RLN = reverse(NLR). Not sure if this is purely coincidental or not. Can anybody either provide a proof of either conjecture? Furthermore, if true, do these conjectures extend to any arbitrary graph/digraph?

This can be proven by induction on trees. I give details on the conjecture 1 here.

• It is clearly true for the empty tree and for leaves;

• Suppose it is true for trees $l$ and $r$. Consider $t$ a node with $l$ and $r$ as left and right children, and $x$ as root. Then, if we denote $Rpre(t)$ the reverse pre-order of $t$, $post(t)$ the post-order of $t$, and $\overline{seq}$ the reversed sequence of $seq$, we get:

$$Rpre(t) = x\cdot Rpre(r)\cdot Rpre(l) = x\cdot \overline{post(r)}\cdot \overline{post(l)} = \overline{post(l)\cdot post(r)\cdot x} = \overline{post(t)}$$

The conjecture 2 is proven in the same way.

I doubt there is an equivalent property for graphs because children are unordered in graphs (but I may be wrong…)

StackExchange Computer Science Q#151687, answer score: 6