Travelling salesman problem? Can't you get to within a factor of 2 of optimal by constructing a minimum spanning tree:
Once you have a MST (which can be built efficiently), the total weight of the tree is a lower bound for the total distance of a TSP solution. However, you can construct a TSP path that traverses each edge of the MST twice; which means that we know that this path (which is easy to find) is at most twice the total cost of the optimal path.
Your solution works only in metric spaces. In non-metric spaces (distances are arbitrary and you are forced to return a cycle, not a tour that repeats vertices) no constant-factor approximation is possible.
Travelling salesman problem? Can't you get to within a factor of 2 of optimal by constructing a minimum spanning tree:
Once you have a MST (which can be built efficiently), the total weight of the tree is a lower bound for the total distance of a TSP solution. However, you can construct a TSP path that traverses each edge of the MST twice; which means that we know that this path (which is easy to find) is at most twice the total cost of the optimal path.