Introduction
I have no school right now, and there's no tutorials on Tree Isomorphism on Codeforces, so I decided to write one :).
What is a tree isomorphism?
Maybe you recognize the word isomorphic: in the context of abstract algebra when two objects are effectively the same, but are labelled differently. We can extend this to trees as well: two trees are isomorphic if they are the same, but may have different node labels. For example,
are isomorphic because we can relabel the nodes. If you want a more rigorous definition, then two trees $$$T_1 = (V_1, E_1)$$$ and $$$T_2 = (V_2, E_2)$$$ are isomorphic iff there exists a $$$\phi$$$ for which $$$\phi(E_1) = E_2$$$ and a $$$\phi^{-1}$$$ for which $$$\phi^{-1} (E_2) = E_1$$$.
What is the problem?
We want to determine if two unrooted trees are isomorphic in $$$\mathcal{O}(N \log N)$$$ time.
If we can solve the problem for rooted trees, can we solve it for unrooted trees?
Yes. We can transform our original problem of checking if two trees are isomorphic into checking if two rooted trees are isomorphic easily. How? We can find the centroids of our first tree and our centroids of our second tree, and then root them at their centroids. So now, we have transformed our unrooted case into a rooted case.
Heuristics
You can skip this, if you'd like. I'm just going to list out some ideas for tree isomorphism checking. All of the following ideas won't solve the problem, but some of them work decently well.
Idea 1: Find the degrees of all nodes and check if $$$\text{oc}_{T1}[i] = \text{oc}_{T2}[i]$$$, where $$$\text{oc}_{T}[i]$$$ is the number of vertices of degree $$$i$$$. If $$$\text{oc}_{T1}[i] = \text{oc}_{T2}[i] \forall i$$$ does not hold, then we know that the trees are not isomorphic.
Idea 2: Root the trees at their centroids. $$$\text{oc}_{T1}[i] = \text{oc}_{T2}[i]$$$, where $$$\text{oc}_{T}[i]$$$ is the number of vertices with subtree size $$$i$$$. If $$$\text{oc}_{T1}[i] = \text{oc}_{T2}[i] \forall i$$$ does not hold, then we know that the trees are not isomorphic.
