# Why is the k-bounded spanning tree problem NP-complete?

The $k$-bounded spanning tree problem is where you have an undirected graph $G(V,E)$ and you have to decide whether or not it has a spanning tree such that each vertex has a degree of at most $k$.

I realize that for the case $k=2$, this is the Hamiltonian path problem. However I'm having trouble with cases where $k>2$. I tried thinking about it in the sense that you can add more nodes onto an existing spanning tree where $k=2$ and maybe since the base is NP complete, adding things on will make it NP-complete as well, but that doesn't seem right. I'm self-studying CS and am having trouble with theory, so any help will be appreciated!

The question has been asked before on stackoverflow, where it has also been answered. The idea is to connect each vertex to $k-2$ new vertices. The new graph has a $k$-bounded spanning tree iff the original graph has a hamiltonian path.
Mohit Singh and Lap Chi Lau gave a polytime algorithm which find a $(k+1)$-bounded spanning tree if a $k$-bounded spanning tree exists. So we can determine the minimum degree of a spanning tree up to an uncertainty of $1$.