An $\textit{articulation point}$ in a connected graph is a vertex such that removing the vertex and its incident edges disconnects the graph into two or more connected components. Let $T$ be a DFS tree obtained by doing DFS in a connected undirected graph $G$. Which of the following options is/are correct?

Let $G$ be a connected undirected graph and $T$ be its DFS tree:

• Option A is incorrect and Option B is correct: By the standard DFS articulation point theorem, the root of a DFS tree $T$ is an articulation point in $G$ if and only if it has two or more children in $T$. If it has only one child, removing it leaves the rest of the graph connected because all other vertices are descendants of that single child and are connected via tree edges and back edges.

• Option C is incorrect: A leaf vertex in a DFS tree has no descendants. Removing a leaf vertex cannot disconnect any remaining vertices in $G$ because all other vertices remain connected via the tree edges of $T$. Hence, a leaf can never be an articulation point.

• Option D is incorrect: If $u$ is an articulation point, it disconnects at least one subtree of its descendants from the rest of the graph. However, it is not necessarily true that *all* descendants $y$ are disconnected from an ancestor $x$, because another descendant $y^{\prime}$ might have a back edge to an ancestor of $u$, providing a path between $x$ and $y^{\prime}$ that does not pass through $u$.

Therefore, only B is correct.