Consider the following saturated chain in the Young's Lattice: $$ \phi \subset \lambda^{1}\subset \lambda^{2}\subset \cdots \subset\lambda^{n}$$ where $\phi$ is the empty partition and $\lambda^{i}$ is a partition of $i$, denoted $\lambda^{i} \vdash i$, for $i=1,2,\ldots ,n$.

I am going to consider the Ferrers diagram of partitions in French notation, and reference the cells of the Ferrers diagram by the pair (row,column).

With the aforementioned saturated chain, associate a column growth word $w=w_1w_2\ldots w_n$ in the following manner: $$w_i= \text{ column of the cell } \lambda^{i}/\lambda^{i-1}$$ Starting from the empty tableau, do jeu-de-taquin slides starting from the outer corner as given by the word $w$ to recover a standard Young tableau(SYT) of shape $\lambda^{n}$. This gives one a(nother) bijection between saturated chains in Young's Lattice and SYTs.

$\textbf{Question}$: Is it possible to predict what the SYT associated with a saturated chain will be, just by the knowing the column growth word, so as to avoid making $n$ jeu-de-taquin slides?

What follows is what I have been doing. I would like a reference where this has already been dealt with, or is implicit (since I am pretty sure this has to be in the literature somewhere).

1) Reverse the word $w$ to obtain $w^r$.

2) Standardise the word $w^r$ to obtain a permutation $\pi$.

3) $\pi \longleftrightarrow (P,Q)$ via the RSK correspondence.

4) $Q^{t}$ (the transpose of the tableau, that is) is the SYT we are looking for.

Example for making the notation clear: Consider the maximal chain $\phi \subset (1)\subset (1,1)\subset (2,1)\subset (2,1,1)\subset (2,2,1)$. Then the column growth word associated with this chain is $w=11212$. Then $w^r=21211$ and the standardisation gives the permutation $41523$ in single line notation.