I don't understand how this algorithm will tell me if a graph is biconnected

Question

I'm doing some practice for an upcoming interview, and a practice question I found asks for an O(V+E) algorithm to tell if a graph is biconnected. This page from Princeton says that a graph is biconnected if it has no articulation vertices, where an articulation vertex is a vertex whose removal will increase the number of connected components (since a biconnected graph should have one connected component).

One common solution to this problem is to do DFS with additional tracking to see if any of the vertices are articulation vertices. This page says that a vertex is an articulation vertex if

• V is a root node with at least 2 children OR
• V has a child that cannot reach an ancestor of V

However, the condition for root nodes doesn't make sense to me. Here is an example of a biconnected graph:

If we choose any of the nodes to be the root, they ALL have 2 or more children, so would be an articulation vertex thus making the graph not biconnected. This is a common algorithm for finding connected components, so assume I have misunderstood something. What do I actually need to check for the root node to see if a graph is biconnected?

Answer 1

You're supposed to do depth-first search, which means that the DFS tree always looks like

1   4
|   |
2---3

because you can't explore the 1--4 link until you're done exploring everything that can be reached from 2 without going through 1, and you won't add 1--4 because it makes a cycle with the tree edges. No node in this tree has two children (1 is the root).