After some searching, I came across exactly what I wanted - seems like knowledge of terms is important. Source from Wikipedia - Biconnected components.
Linear time depth first search
The classic sequential algorithm for computing biconnected components in a connected undirected graph is due to John Hopcroft and Robert Tarjan (1973). It runs in linear time, and is based on depth-first search. This algorithm is also outlined as Problem 22-2 of Introduction to Algorithms (both 2nd and 3rd editions).
The idea is to run a depth-first search while maintaining the following information:
the depth of each vertex in the depth-first-search tree (once it gets visited), and
for each vertex v, the lowest depth of neighbors of all descendants of v in the depth-first-search tree, called the lowpoint.
The depth is standard to maintain during a depth-first search. The lowpoint of v can be computed after visiting all descendants of v (i.e., just before v gets popped off the depth-first-search stack) as the minimum of the depth of v, the depth of all neighbors of v (other than the parent of v in the depth-first-search tree) and the lowpoint of all children of v in the depth-first-search tree.
The key fact is that a nonroot vertex v is a cut vertex (or articulation point) separating two biconnected components if and only if there is a child y of v such that lowpoint(y) ≥ depth(v). This property can be tested once the depth-first search returned from every child of v (i.e., just before v gets popped off the depth-first-search stack), and if true, v separates the graph into different biconnected components. This can be represented by computing one biconnected component out of every such y (a component which contains y will contain the subtree of y, plus v), and then erasing the subtree of y from the tree.
The root vertex must be handled separately: it is a cut vertex if and only if it has at least two INDEPENDENT (non-connectible) children. However, since we hold the child count of each visited node, and the count only increments if the adjacent node being checked is not already visited, the root having >=2 children count would surely mean that at least two of the root's children are independent. If there were no 2 independent children of the root, all of the children of the root apart from the first checked would be marked as visited BEFORE the DFS search returned to root and continued the search to the second child, therefore the root's children count would be stuck at 1.
The code below prints the said articulation points, but with minimal changes it can also just print out the count of the said points.
visited[i] = true
depth[i] = d
low[i] = d
childCount = 0
isArticulation = false
for each ni in adj[i]
if not visited[ni]
parent[ni] = i
GetArticulationPoints(ni, d + 1)
childCount = childCount + 1
if low[ni] >= depth[i]
isArticulation = true
low[i] = Min(low[i], low[ni])
else if ni <> parent[i]
low[i] = Min(low[i], depth[ni])
if (parent[i] <> null and isArticulation) or (parent[i] == null and childCount > 1)
Output i as articulation point