Breadth first search is complete even if the branches are infinitely deep. In the sense that, if there is a solution it will find it eventually.
For example, each node has unique path to root, so write <n1, n2, ..., nk> where each ni is the sibling ordinal of the node at depth i in that path, i.e. it's the ni-th sibling of the n(i-1)st node. Raising each of these to the ith prime and taking a product gives each node a unique integer label. Traverse nodes in label order and voilà?
However, that all assumes we know the tree beforehand, which doesn't make sense for generic call trees. Do we just smash headfirst into Rice on this when trying to traverse in complete generality?
Breadth first search would visit every node breadth first, so given infinite time, the solution would eventually be visited.
Meanwhile, say a branch had a cycle in it, even given infinite time, a naive depth first search would be trapped there, and the solution would never be found.
Or, suppose that a node has infinitely many children, but the first child has its own child. A BFS would get stuck going through all the first-level children and never reach the second-level child.
A BFS-like approach could work for completeness, but you'd have to put lower-level children on the same footing as newly-discovered higher-level children. E.g., by breaking up each list of children into additional nodes so that it has branching factor 2 (and possibly infinite depth).
The Wikipedia article is fairly useful: https://en.wikipedia.org/wiki/Countable_set