We can rephrase this problem in terms of graph theory, whereby we need to transform a rooted tree by disconnecting and reconnecting the tree. Refer to the initial tree as ~A~, and the desired tree as ~B~. Clearly, the root of the tree cannot change, so if ~A~ and ~B~ have different roots, then no solution exists. Otherwise, a solution exists, which can be proved through the following construction.

Each move updates the parent of at most one node, which lower bounds the answer at the number of mismatches between ~S~ and ~T~. In fact, the following construction shows that this lower bound is actually achievable.

We process nodes by the DFS ordering of ~B~. When we process a node ~X~, we update its parent to ~T_X~ if necessary. This works because when we process node ~X~, then all of the ancestors of node ~X~ in ~B~ will already be in the correct position in the current tree. Actually, a BFS ordering of ~B~ works for the same reason. In fact, if we consider edges in ~B~ to be directed away from the root, then any topological ordering of ~B~ suffices.

As a final note, note that reversing the order of the moves yields a valid reorganisation of the tree from ~B~ to ~A~. This implies that we could instead direct the edges in ~A~ towards the root, and process the nodes in any topological ordering.

Time complexity: ~\mathcal{O}(N)~