The Littelmann path model defines $e$$e_i$ and $f$$f_i$ operators which correspond in type A to the $e$$e_i$ and $f$$f_i$ operators on semistandard Young tableaux (i.e. since they both are different ways of looking at Type A crystal.)
In the context of semistandard Young tableaux, we can think of the $e$$e_i$ and $f$$f_i$ operators as being described by a paired parenthesis operation, and that operation commutes with RSK insertion thus the $e$$e_i$ and $f$$f_i$ operators are often thought of as acting on words, not just semistandard Young tableaux.
In the context of the Littelmann path model, there are many paths to the same point, all equally compatible with the same $e$$e_i$ and $f$$f_i$ operators. Some of those paths, particularly those that are a sequence of various length one steps from the set $\{e_i-v\}_{i=1\cdots,n}$ (where $v=\frac{1}{n}(1,1,\cdots,1)$ so the vector sums to zero) might reasonably be considered an analogue of a word, with $e_i- v$ corresponding to i in the word (with the end of the word corresponding to the start of the path at the origin). Call these path "special" paths for purposes of this post.
Since every highest weight path of the same weight generators the same crystal, there's a sense in which order doesn't matter to the $e$$e_i$ and $f$$f_i$ operators in the Littelmann path model. In contrast, not every word of the same type RSK inserts to the same semistandard Young Tableaux, so special paths must not be fully compatible with RSK and the $e$$e_i$ and $f$$f_i$ paired parenthesis operators on words. Is there a natural subset of special paths that is consistent on which the paired parenthesis $e$$e_i$ and $f$$f_i$ operators are consistent with the $e$$e_i$ and $f$$f_i$ operators of the Littelmann path model? For example, on special paths which correspond to reading words of Semistandard Young tableaux, will the $e$$e_i$ and $f$$f_i$ operators in the Littelmann path model result in paths which are also the reading word of semistandard Young tableaux?
I suspect this is too much to ask for, although it seemed to work on a couple of very small examples.