  The three-in-a-tree algorithm of Chudnovsky and Seymour decides in time
$O(n^4)$ whether three given vertices of a graph belong to an induced tree.
Here, we study four-in-a-tree for triangle-free graphs. We give a structural
answer to the following question: what does a triangle-free graph look like if
no induced tree covers four given vertices? Our main result says that any such
graph must have the "same structure", in a sense to be defined precisely, as a
square or a cube.
  We provide an $O(nm)$-time algorithm that given a triangle-free graph $G$
together with four vertices outputs either an induced tree that contains them
or a partition of $V(G)$ certifying that no such tree exists. We prove that the
problem of deciding whether there exists a tree $T$ covering the four vertices
such that at most one vertex of $T$ has degree at least 3 is NP-complete.
