Understanding bounded-diameter decomposition of graphs for PTAS While trying to understand Baker's approach (also explained in this article by Eppstein) to designing PTAS's for planar graphs, I came across a difficulty. 

The idea is, given an integer $k$, decompose the graph into subgraphs of tree-width bounded by $O(k)$ (which for planar graphs happens if $diameter = O(k)$), then solve the problem on the subgraphs in polynomial-time, and finally glue the solutions together to get a polynomial-time $(1 + O(1/k))$-approx.

The decomposition method described in the part $8$ of the article, amounts to, as far as I understood, pick any vertex, get a BFS tree from that vertex, and remove every $k$-th level of the tree (potentially starting at level $0 \leq i &lt; k$, but I do not think it matters for this part).

The author then argues that the obtained subgraphs (consisting each of at most $k-1$ levels of the BFS tree), have bounded diameter (i.e. depending only on $k$). 

This is the part I do not understand. For instance, what happens if some layer of the tree is a long path ? How is the diameter of a subgraph containing this layer bounded ?

Thank you for any help,

Lucas
Actually, the obtained subgraphs may not have bounded diameter, but they still have bounded tree-width. In order to see this, take one such subgraph $H$, and build $H'$ by adding a new vertex $u$ that is connected to each of the top-level vertices of $H$ (remember that $H$ was obtained as $k-1$ consecutive layers of a BFS tree). Then clearly $diam(H') \leq 2k$ since any vertex in $H'$ is at distance less than $k$ of $u$. We then know that $tw(H') = O(diam(H')) = O(k)$. Finally, $H$ is a minor of $H'$ so $tw(H) \leq tw(H')$, hence $tw(H) = O(k)$.