Root the tree arbitrarily. Let sz[v] be the size of the subtree of node v, including it (to make the implementation easier).

The answer for query v is max(sz[c] for all children c of v, n-sz[v]), because cutting off node v gives you a tree for each child and another for the rest of nodes. And you can precompute this to not get $$$O(NQ)$$$ runtime.

Total complexity: $$$O(N + Q)$$$

If nodes get removed after each query, then I think you'd need Euler Tour Technique + Segment Tree w/ Lazy Prop to subtract from all nodes in the subtree of v, and maybe something else for the ancestors of v, which would be a lot more complicated. I'm not too sure about this though, so maybe someone knows a better (and simpler!) way of doing this.

Approach 1. Compute the low function on the graph's dfs tree (https://cp-algorithms.com/graph/cutpoints.html similar to this). When removing node V, the subtree of V's child will be disconnected from the rest of the graph if low[child] > depth[V]. The largest connected component will be equal to Max(disconnected_subtree_sizes, N-sum_of_disconnected_subtree_sizes). You can compute the answer for each node offline, and answer in O(1) for a total of O(n+m).

Approach 2. You can solve the problem using dynamic connectivity (https://cp-algorithms.com/data_structures/deleting_in_log_n.html) in O((n+m)log(m)log(n)) by simulating removing all neighbouring edges for each node at a time while also keeping a disjoint set union data structure.