It gives a set-theoretic bijection, but not one that plays by all the rules I mentioned above. In particular, you don't get a bijection that corresponds to a finite, non-looping program.
Generally speaking, Cantor–Schröder–Bernstein does not give you a (finite) construction of the bijection, because you have to follow the inverses of f and g back until you either end up with something from the first set with no preimage under g, or something from the second set with no preimage under f, or you find that you are iterating forever. You have to make a decision based on whether or not a certain computation terminates. That's essentially why it is a non-constructive theorem (in fact, it even implies the law of excluded middle)
But in your specific case, we're kind of in luck. The injections are simple enough that you can work out the bijection that you'd get from König's proof manually, and it goes like this:
Concretely, let's write (AB) for the tree with left subtree A and right subtree B, and write . for the empty tree. Your injections are f(A,B) = (AB) and g(T) = (T.). Alternately applying f/g partitions the set of trees into a bunch of infinite sequences. Your bijection is given by: if T is part of the sequence that begins with the empty tree, pair T with (T,.). If T is part of some other sequence, then T is not the empty tree; it is of the form T = (LR), and you should pair T with (L,R). Concretely, the sequence that starts with the empty tree looks like:
In other words, the single trees that appear in the empty list's sequence are the fully-left-leaning trees like ((((..).).).); all other trees are in the other sequences. So to decide where your tree goes in the bijection, you have to do this:
if (T is fully left-leaning) then (T,.) else (left-child(T), right-child(T))
And computing whether or not T is fully left-leaning involves an unbounded amount of computation. You have to actually walk the whole tree. So this bijection won't correspond to a finite, non-looping program. In a sense, the algorithm you get from Cantor–Schröder–Bernstein is not "continuous", but the one you get from the Seven Trees In One construction is.
Generally speaking, Cantor–Schröder–Bernstein does not give you a (finite) construction of the bijection, because you have to follow the inverses of f and g back until you either end up with something from the first set with no preimage under g, or something from the second set with no preimage under f, or you find that you are iterating forever. You have to make a decision based on whether or not a certain computation terminates. That's essentially why it is a non-constructive theorem (in fact, it even implies the law of excluded middle)
But in your specific case, we're kind of in luck. The injections are simple enough that you can work out the bijection that you'd get from König's proof manually, and it goes like this:
Concretely, let's write (AB) for the tree with left subtree A and right subtree B, and write . for the empty tree. Your injections are f(A,B) = (AB) and g(T) = (T.). Alternately applying f/g partitions the set of trees into a bunch of infinite sequences. Your bijection is given by: if T is part of the sequence that begins with the empty tree, pair T with (T,.). If T is part of some other sequence, then T is not the empty tree; it is of the form T = (LR), and you should pair T with (L,R). Concretely, the sequence that starts with the empty tree looks like:
. -> . . -> (..) -> (..) . -> ((..).) -> ((..).) . -> (((..).).) -> **
In other words, the single trees that appear in the empty list's sequence are the fully-left-leaning trees like ((((..).).).); all other trees are in the other sequences. So to decide where your tree goes in the bijection, you have to do this:
if (T is fully left-leaning) then (T,.) else (left-child(T), right-child(T))
And computing whether or not T is fully left-leaning involves an unbounded amount of computation. You have to actually walk the whole tree. So this bijection won't correspond to a finite, non-looping program. In a sense, the algorithm you get from Cantor–Schröder–Bernstein is not "continuous", but the one you get from the Seven Trees In One construction is.