Merge two trees on equal nodes - algorithm

Question

Nodes are equal when their IDs are equal. IDs in one tree are unique. On the schemas, IDs of nodes are visible.
Consider tree1:
root
|
+-- CC
| |
| \-- ZZZ
| |
| \-- UU
|
\-- A
|
\-- HAH
And tree2:
root
|
+-- A
|
+-- ADD
|
\-- HAH
I would like that merge(tree1, tree2) will give this:
root
|
+-- CC
| |
| \-- ZZZ
| |
| \-- UU
|
\-- A
|
+-- HAH
|
\-- ADD
How to do it?
Node has typical methods like getParent(), getChildren().
Order of the children doesn't matter. So, the result could be also:
root
|
+-- A
| |
| +-- ADD
| |
| \-- HAH
|
\-- CC
|
\-- ZZZ
|
\-- UU

Answer 1

My proposition in pseudocode. Comments are more than welcome.
merge(tree1, tree2) {
for (node : tree2.bfs()) { // for every node in breadth-first traversal order
found = tree1.find(node.getParent()); // find parent in tree1
if (found == null) // no parent?
continue; // skip it, it's root
if (!found.getChildren().contains(node)) // no node from tree2 in tree1?
found.add(node); // add it
}
return tree1;
}

Answer 2

The basic algorithm is not hard:
def merge_trees (t1, t2):
make_tree(map(merge_trees,assign(getChildren(t1),getChildren(t2),tree_similarity)))
make_tree(children): create a tree with the given list of children
map(f,list): calls function f on each element of list and return the list of return values
assign(list1,list2,cost_function): implements the Hungarian algorithm, returning the list of matched pairs
The trick is in defining tree_similarity which would have to call assign recursively.
In fact, the efficient implementation would have to cache the return values of the assign calls.