Longer answer: actually another common feature is that both can execute code at compile-time (macros).
Full answer: really, no. They are fundamentally different: Lisp is a decent symbolic language, Forth is a decent portable assembler.
In Lisp we know that (eql 1 2 3) is a "too many arguments" error. It doesn't just take 1 2 from the stack, push back the answer, and leave 3 beneath it.
Lisps are actually fairly conventional block-structured languages, having more in common with Pascal than with Forth. Lexical scopes, local variables, function calls with arguments going to formal parameters and all that.
That error also comes from the semantics. You can't tell that (eql 1 2 3) is a too-many-arguments error just by looking at the tokens. That could be a 3-ary function, or an "eql is not a function" error.
The parentheses in Lisp code certainly suggest a particular tree structure, but they don't require it by virtue of their graphical form. And once you're willing to look up the arity of functions, Forth is written directly in its own tree structure to exactly the same degree that Lisp is, just with structural markers elided.
Think about it another way: a classmate in compilers class asked me once how Lisp handled operator precedence. Obviously, with the tree structure of the code fully specified, Lisp doesn't handle operator precedence because the question can't arise.
Note that this is just as true of Forth.
Also, Lisp supports variadic functions with the same safety. Delimitation of arguments is necessary for variadic functions, like list.
The structure of every top-level form is an actual tree object in memory that is constructed before anything else happens. It can be walked by macros, or quoted so that the program accesses it as data directly.
Basically, nothing like Forth.
This is not the case; you can easily have a Forth word that consumes however many arguments as are on the stack.
Is there more to the relationship than "Lisp is written as a preorder tree, and Forth is written as a postorder tree"?