If we suppose that a graph is of a tree topology (connected, acyclic graph), does there exist a straight forward approach (algorithm/heuristic) for solving the k-Chinese Postman Problem (aka the k-Route Inspection Problem)?
For example, if we pick one vertex to be the starting point, the solution for k=1 postman and all edge distances equal to 1 would just be the solution to the Chinese Postman Problem on a tree, which is just 2(|V|-1), with each edge being traversed exactly twice. If k>1, the edges traversed by each postman would need to be split in some way.