Consider the following variation of the Chinese postman problem (also known as the route inspection problem). Instead of finding the shortest closed walk which traverses each edge at least once, find the shortest open walk which traverses each edge at least once.
This variation is sometimes more interesting. For example, the length of a minimal closed edge-covering walk on every tree on $n$ vertices is $2(n-1)$ (as every bridge must be crossed twice), but the length of a minimal open edge-covering walk is different for different trees on the same vertex set. For example, that number is $6$ for a path with $6$ edges, and $10$ for a star with $6$ leaves.
Does this problem have an easy solution, or an efficient algorithm, as the Chinese postman problem?
I'll add that I am not really interested in finding a shortest walk, but rather in figuring out the length of a minimal walk.