Let $G$ be a simple undirected graph, $T_D$ be a DFS tree on $G$, and $T_B$ be the BFS tree on $G$.

Consider the following statements.

Statement I: No edge of $G$ is a cross with respect to $T_D$.

Statement II: For every edge $(u,v)$ of $G$, if $u$ is at depth $i$ and $v$ is at depth $j$ in $T_B$, then $|i-j|=1$.

In the light of the above statements, choose the correct answer from the options given below:

Statement I is true.

In DFS of an undirected graph, there are no cross edges.

Edges are generally tree edges or back edges.

So, no edge of $G$ is a cross edge with respect to DFS tree $T_D$.

Statement II is false.

In BFS tree, for every edge $(u,v)$, the depths of $u$ and $v$ differ by at most $1$.

So, the correct condition is:

$|i-j|\le 1$

But the statement says:

$|i-j|=1$

This is not always true because two adjacent vertices can also be at the same level in a BFS tree.

Therefore, Statement I is true and Statement II is false.