Let $G$ be a connected graph and $v$ be a vertex in $G$. Suppose a DFS traversal from $u$ is performed resulting in a tree $T$, and a BFS from $u$ also results in the same tree $T$. I would like to show that $T = G$.
My main problem for this is that I am not sure how to initially approach this proof, and given my current approach is within reason I am stuck as for my next step.
My current thought is to go by contradiction and suppose that $T$ is both the DFS and BFS tree of $G$ with $u$ being the root, but $T \neq G$. As $T \subset G$, there must exist an edge $e \in G$ such that $e \notin T$ (since $G$ is connected it cannot be the case that there is a vertex in $G$ that is not in $T$). From here I believe my contradiction will come from the DFS and BFS traversals not being equivalent, but this is where I am stuck in this approach.
Could anyone hint at my next step or give me an idea for another approach to the proof?