If you use a Maximum Likelihood method, you will get a "score" of how good the "best" tree is. For example, the best tree might have a Likelihood score of -2000.0. This tree (T0) might group (A,B) together, (C,D) together, with (E) as an outgroup. Therefore the treefile, without branchlengths, would look like this: ((A,B),(C,D), E);
You could ask if another tree (T1) was significantly worse than the "best" tree. For example, consider tree ((A,C),(B,D),E);. This will have a poorer Likelihood (simply because it is not the "best" tree) so maybe something like -2010.0.
We can do a statistical test to see if Tree T1 is significantly worse than Tree T0 i.e. is Likelihood difference of 10 units significantly greater than a difference of zero.. We can use the Shimodaira-Hasegawa test to do this e.g. with the package IQ-TREE, easily accessed online via the W-IQ-TREE server. You would provide the alignment and at the bottom of the webpage, provide the 2 treefiles concatenated:
((A,B),(C,D), E);
((A,C),(B,D),E);
If you compare more than 2 trees, the "approximately unbiased" test (UA) is perhaps better. This can also be done using IQ-TREE.
If you HAVE NOT done a phylogenetic analysis - and thus do not know the "best" tree in advance - then you can provide an alignment and a concatenated list of treefiles and then carry out the Hasegawa-Kishino test. This HK statistical test is a two-sided test, unlike SH and UA which are one-sided statistical tests. Again available in IQ-TREE.
These tests can be done in other Maximum Likelihood packages (e.g. RAxMl, PhyML). IQ-TREE is just an example.
This is a follow up question: How to optimize Matrix by Differential evolution algorithm?? that might explain where OP is going. However, I am still struggling with finding out what the aim is.