Boxed Like a Fish

Winning vertices

Call a vertex good if Cyndaquil can force reaching a leaf from there. Obviously, every leaf is good.

The main observation is a closure rule:

$$\text{if } a,b \text{ are good and } \operatorname{dist}(a,b) \le k+1, \text{ then every vertex on } a\leadsto b \text{ is good.}$$

Why? Stand at some vertex $x$ on that path. Snorlax can block at most one edge, so he cannot block both directions toward $a$ and toward $b$. Cyndaquil moves toward an unblocked side. If Snorlax has just reset the cooldown, then from any internal point the chosen good endpoint is at distance at most $k$, so Cyndaquil reaches it before Snorlax can change the blocked edge again. If Snorlax waits and reacts one move later, Cyndaquil either already reached the chosen endpoint, or can switch to the other endpoint within $k$ moves. Then we use the strategy from that good endpoint. One blocked edge is not a wall; it is duct tape.

Why this condition is also necessary

Let $W$ be the smallest set containing all leaves and closed under the rule above. If Cyndaquil starts outside $W$, Snorlax can keep him outside forever.

Look at the connected component of $T\setminus W$ containing Cyndaquil. It has no leaves, since all leaves are in $W$. Snorlax waits with cooldown $0$ until Cyndaquil is about to cross an edge from this component into $W$, then blocks exactly that edge.

Suppose two boundary exits are $x-a$ and $y-b$, where $x,y$ are outside $W$ and $a,b\in W$. If

$$\operatorname{dist}(x,y)<k,$$

then

$$\operatorname{dist}(a,b)=\operatorname{dist}(x,y)+2\le k+1,$$

so the closure rule would have put the whole path, including $x$ and $y$, into $W$. Contradiction. Therefore different exits are at least $k$ moves apart inside the bad component. After Snorlax blocks one exit and resets cooldown to $k$, Cyndaquil needs at least $k$ moves just to reach another boundary vertex, and one more move to cross; by then Snorlax is ready again. So outside $W$ is a real jail, not a skill issue.

Therefore the answer is YES exactly when $v\in W$.

Tree DP

Root the tree at $v$. For each node $x$, compute:

$$dp[x]= \begin{cases} 0, & x \text{ is good inside its rooted subtree},\\ \text{distance from }x\text{ to the closest good vertex in its subtree}, & \text{otherwise}. \end{cases}$$

A rooted leaf has $dp[x]=0$.

For an internal node $x$, collect the $dp$ values of its children. Let the two smallest be $a\le b$.

If there are two child subtrees, the closest good vertices from those two sides have distance

$$(a+1)+(b+1)=a+b+2$$

from each other through $x$. So $x$ becomes good iff

$$a+b+2\le k+1 \iff a+b<k.$$

If this happens, set $dp[x]=0$. Otherwise, the closest good vertex below $x$ is through the best child, so

$$dp[x]=a+1.$$

If $x$ has only one child, no pair can be formed through $x$, so the same formula $dp[x]=a+1$ applies.

Finally, print YES iff

$$dp[v]=0.$$

The DP is linear:

$$O(n)$$

per test case, and

$$O\left(\sum n\right)$$

over the whole input. I also checked the recurrence against the official Codeforces tutorial: https://codeforces.com/blog/entry/151886.