probably it can be (over)solved by root rotating and mex on segment queries.
supporse we know mex for each subtree initially (root=0). if r is root and c its child we can rotate this edge to make c as root. how will mexes change? mex(c) will be mex of all numbers and mex(r) will be mex of numbers except subtree of c (current subtree)
so current subtree is defined by oriented edge. in tree with root 0 this will be subtree or... undertree (how to call it better?). if we concate euler tour of tree twice than both subtree and undertree becomes segment. and we can ask mex on segment in O(logn).
upd (I got AC). ok we need only mex of each subtree (for init) and mex for each not subtree (for rotating). this can be found somehow easy or by mex on segments query (O(qlogn)).
probably it can be (over)solved by root rotating and mex on segment queries.
supporse we know mex for each subtree initially (root=0). if r is root and c its child we can rotate this edge to make c as root. how will mexes change? mex(c) will be mex of all numbers and mex(r) will be mex of numbers except subtree of c (current subtree)
so current subtree is defined by oriented edge. in tree with root 0 this will be subtree or... undertree (how to call it better?). if we concate euler tour of tree twice than both subtree and undertree becomes segment. and we can ask mex on segment in O(logn).
upd (I got AC). ok we need only mex of each subtree (for init) and mex for each not subtree (for rotating). this can be found somehow easy or by mex on segments query (O(qlogn)).