The rendezvous task calls for two mobile agents, starting from different nodes of a network modeled as a graph to meet at the same node. Agents have different labels which are integers from a set $\{1,\dots,L\}$. They wake up at possibly different times and move in synchronous rounds. In each round, an agent can either stay idle or move to an adjacent node. We consider deterministic rendezvous algorithms. The time of such an algorithm is the number of rounds since the wakeup of the earlier agent till the meeting. In this paper we consider rendezvous in infinite trees. Our main goal is to study the impact of orientation of a tree on the time of rendezvous. We first design a rendezvous algorithm working for unoriented regular trees, whose time is in $O(z(D) \log L)$, where $z(D)$ is the size of the ball of radius $D$, i.e, the number of nodes at distance at most $D$ from a given node. The algorithm works for arbitrary delay between waking times of agents and does not require any initial information about parameters $L$ or $D$. Its disadvantage is its complexity: $z(D)$ is exponential in $D$ for any degree $d>2$ of the tree. We prove that this high complexity is inevitable: $Ω(z(D))$ turns out to be a lower bound on rendezvous time in unoriented regular trees, even for simultaneous start and even when agents know $L$ and $D$. Then we turn attention to oriented trees. While for arbitrary delay between waking times of agents the lower bound $Ω(z(D))$ still holds, for simultaneous start the time of rendezvous can be dramatically shortened. We show that if agents know either a polynomial upper bound on $L$ or a linear upper bound on $D$, then rendezvous can be accomplished in oriented trees in time $O(D\log L)$, which is optimal. When no such extra knowledge is available, we design an algorithm working in time $O(D^2+\log ^2L)$.