The randomized query complexity of the conjoined trees problem
I think I have a deterministic algorithm that finds the exit in $O(n2^{n/2})$ oracle calls.

First, find the labels for all the vertices of distance $n/2$ from the entrance. This takes $O(2^{n/2})$ queries. Then, start from the entrance and walk $n+1$ steps to get to a node $X$ of distance $n+1$ from the entrance. We will try to reach the exit from this node.

We have two options of where to go from $X$, and we wish to choose the one closer to the exit. To do this, pick one of the options arbitrarily, arriving at a node $Y$. Then explore all the $O(2^{n/2})$ ways of walking $n/2$ steps from $Y$. If one of them gives a label of distance $n/2$ from the entrance, we know that going from $X$ to $Y$ was the wrong choice. Otherwise, $Y$ was the right choice. Thus in $O(2^{n/2})$ we found a node $X_2$ of distance $n+2$ from the entrance.

We can keep proceeding this way. To find a node of distance $n+3$ from the entrance, we start at $X_2$ and walk two arbitrary steps. We then explore all options of walking $n/2$ additional steps (while never walking backwards), and check if any of them have distance $n/2$ from the entrance. This will happen if and only if the first step we walked from $X_2$ was wrong.

Getting to the exit requires doing this $n$ times, for a total of $O(n2^{n/2})$ oracle calls. In addition, perhaps surprisingly, this algorithm is deterministic.

Edit: Just to clarify, to get from $X_t$ to $X_{t+1}$, we walk $t$ arbitrary steps, then run a search of depth $n/2$, for a total of $t+2^{n/2}$ steps. If the first step lead away from the exit, then all the first $t$ steps did, and we therefore find a label of distance $n/2$ from the entrance. This means $t+2^{n/2}$ steps suffice for determining the next single step to take from $X_t$.