\section{The general case}\label{sec:general}
In this Section, we prove our main result, \cref{thm:main}, in full generality. Fix a strict partition $\lambda\subset Z(d,r)$ such that $\lambda_d>0$ and set $N = |\lambda|$. Let $\ell = |Z(d,r)| = \ell(w^{(d,r)})$ and set $\mu_0 = \sigma_0 = 0$, $\mu_i = Z(d,r)_i-\lambda_i$ and $\sigma_i = \sum_{k=1}^{i}\mu_k$ for all $i\in [d]$. Recall our main framework \[\SYT(\lambda)\longleftrightarrow\SYT(Z(d,r))|_{\lambda}\longleftrightarrow\Red(w^{(d,r)})|_{\lambda}\longleftrightarrow\BS(Z(d,r))|_{\lambda}\longleftrightarrow\BS(\lambda).\] The first arrow of bijection is immediate (\cref{def:SYT|lambda}). In each of the subsequent subsections, we prove one remaining bijection respectively. A complete example is given at the end of this section in \cref{ex:main-ex}. Readers are encouraged to refer to this example for intuition. 
\subsection{Bijection between \texorpdfstring{$\SYT(Z(d,r))|_{\lambda}$}{} and \texorpdfstring{$\Red(w^{(d,r)})|_{\lambda}$}{}}
\begin{defin}\label{def:SYT|lambda}
    For any tableau $T\in \SYT(\lambda)$, define $T^{+}\in \SYT(Z(d,r))$ to be the tableau obtained from $T$ by assigning $N+1,\ldots,\ell$ to the cells in $Z(d,r)\setminus \lambda$ from left to right along rows and from top to bottom. Define $\SYT(Z(d,r))|_{\lambda}$ to be the set of all such $T^+$ obtained from some $T\in \SYT(\lambda)$.
\end{defin}
\begin{ex}
Let $\lambda = (6,2,1)\subset Z(3,2)$ and 
\[\ytableausetup{boxsize=1.5em}
T = \begin{ytableau}
1 & 2 & 3 & 5 & 6 & 9 \\
\none & 4 & 7 & \none & \none \\
\none & \none & 8 & \none 
\end{ytableau}, \text{ then }
T^{+} = \begin{ytableau}
1 & 2 & 3 & 5 & 6 & 9 & \circled{10}\\
\none & 4 & 7 & \circled{11} & \circled{12} & \circled{13}\\
\none & \none & 8 & \circled{14} & \circled{15}
\end{ytableau}\, .
\]
\end{ex}

\begin{defin}\label{def:alambda}
    Define the word $\mathbf{a}^\lambda= \mathbf{a}_1^\lambda\cdots \mathbf{a}_d^\lambda$ where
    \begin{equation}\label{eqn:Mar21aaa}
        \mathbf{a}_i^\lambda = d+r-i-\mu_i+1,\ldots, d+r-i.
    \end{equation}
    Define $\Red(w^{(d,r)})|_{\lambda}$ to be the set of reduced words ending with $\mathbf{a}^\lambda$.
\end{defin}

\begin{prop}\label{prop:alambda}
The bijection $\mathbf{a}\mapsto Q(\mathbf{a})$ between $\Red(w^{(d,r)})$ and $\SYT(Z(d,r))$ in \cref{cor:uniquePtableau} induces a well-defined bijection between $\Red(w^{(d,r)})|_{\lambda}$ and $\SYT(Z(d,r))|_{\lambda}$.
\end{prop}

A key ingredient in the proof of \cref{prop:alambda} is the existence of the \emph{reverse} of Kra\'skiewicz's insertion.
\begin{lemma}[Lemma~1.25, \cite{Lam1995thesis}]\label{lemma:reverse-insertion}
Given $(P(\mathbf{a}),Q(\mathbf{a}))$ for some $\mathbf{a} = a_1a_2\cdots a_{\ell(w)}\in \Red(w)$ and $w\in W(B_n)$, let $Q'$ be obtained by removing the largest entry in $Q(\mathbf{a})$. Then there is a unique $a\in [0,n-1]$ and a unique $P'\in \SDT(ws_{a})$ such that $P' \leftarrow a = P,$ and $\sh(P') = \sh(Q')$. In fact, we have $a = a_{\ell(w)}$.
\end{lemma}

\begin{comment}
\begin{ex}\label{ex:w-lam}
Consider $\lambda=(6,2,1)\subset Z(3,2)$. Here $w^{(3,2)}=45\bar1\bar2\bar3$. We compute $\mathbf{a}^\lambda$ and $w^{\lambda}$ step by step via (the reverse of) Kra\'skiewicz's insertion. We start with the unique $P^+\in \SDT(w^{(3,2)})$, and a standard shifted tableau $Q^+$ of shape $Z(3,2)$, padded from any standard shifted tableau $Q$ of shape $\lambda$, shown in \cref{tab:ex-w-lam}. 

\begin{table}[h!]
\centering
\setlength{\extrarowheight}{10pt}
\ytableausetup{smalltableaux}
\begin{tabular}{c|c|c}
\text{insertion tableau} $P$ & \text{recording tableau} $Q$ & letter $a_i$'s \\
\begin{ytableau}
4 & 3 & 0 & 1 & 2 & 3 & 4 \\
\none & 3 & 0 & 1 & 2 & 3\\
\none & \none & 0 & 1 & 2
\end{ytableau} & 
\begin{ytableau}
*(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & 10\\
\none & *(yellow) & *(yellow) & 11 & 12 & 13\\
\none & \none & *(yellow) & 14 & 15
\end{ytableau} &  \\
\begin{ytableau}
4 & 2 & 0 & 1 & 2 & 3 & 4 \\
\none & 2 & 0 & 1 & 2 & 3\\
\none & \none & 0 & 1
\end{ytableau} & 
\begin{ytableau}
*(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & 10\\
\none & *(yellow) & *(yellow) & 11 & 12 & 13\\
\none & \none & *(yellow) & 14
\end{ytableau} & $a_{15}=2$ \\
\begin{ytableau}
4 & 1 & 0 & 1 & 2 & 3 & 4 \\
\none & 1 & 0 & 1 & 2 & 3\\
\none & \none & 0
\end{ytableau} & 
\begin{ytableau}
*(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & 10\\
\none & *(yellow) & *(yellow) & 11 & 12 & 13\\
\none & \none & *(yellow)
\end{ytableau} & $a_{14}=1$ \\
\begin{ytableau}
3 & 1 & 0 & 1 & 2 & 3 & 4 \\
\none & 1 & 0 & 1 & 2 \\
\none & \none & 0
\end{ytableau} & 
\begin{ytableau}
*(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & 10\\
\none & *(yellow) & *(yellow) & 11 & 12\\
\none & \none & *(yellow)
\end{ytableau} & $a_{13}=3$ \\
\begin{ytableau}
2 & 1 & 0 & 1 & 2 & 3 & 4 \\
\none & 1 & 0 & 1  \\
\none & \none & 0
\end{ytableau} & 
\begin{ytableau}
*(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & 10\\
\none & *(yellow) & *(yellow) & 11 \\
\none & \none & *(yellow)
\end{ytableau} & $a_{12}=2$ \\
\begin{ytableau}
2 & 1 & 0 & 1 & 2 & 3 & 4 \\
\none & 1 & 0   \\
\none & \none & 0
\end{ytableau} & 
\begin{ytableau}
*(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & 10\\
\none & *(yellow) & *(yellow) \\
\none & \none & *(yellow)
\end{ytableau} & $a_{11}=1$ \\
\begin{ytableau}
2 & 1 & 0 & 1 & 2 & 3 \\
\none & 1 & 0   \\
\none & \none & 0
\end{ytableau} & 
\begin{ytableau}
*(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) & *(yellow) \\
\none & *(yellow) & *(yellow) \\
\none & \none & *(yellow)
\end{ytableau} & $a_{10}=4$ \\
\end{tabular}
\caption{An example of $\mathbf{a}^{\lambda}$ and $w^{\lambda}$}
\label{tab:ex-w-lam}
\end{table}
We read off $\mathbf{a}^{\lambda}=a_{10}\cdots a_{15}=412312$. As $w^{\lambda}=w^{(d,r)}s_{a_{15}}\cdots s_{a_{10}}$, we obtain $w^{\lambda}=\bar2\bar14\bar35$.
\end{ex}

\begin{remark}
Here is an explicit description for $w^\lambda$. Draw a $(d+r)\times(d+r)$ triangle over $\lambda$ with box $(1,1)$ as the top left corner. Rotate by $45^\circ$ and consider the Dyck path of semilength $d+r$ that is the border of $\lambda$. Label the upsteps with $d+r,d+r-1,\dots,d+1,\bar1,\bar2,\dots,\Bar{d}$ in order, and pass these labels to the corresponding leftmost downsteps. Reading off the downsteps in order gives $w^\lambda$. An example is shown in \cref{fig:read-wlambda}.
\begin{figure}[h!]
    \centering
    \begin{tikzpicture}[scale = 0.5]
    \draw (0,0) -- ++(4,4) --++(3,-3) -- ++(1,1) --++(2,-2) -- cycle;
    \draw (2,2) -- ++(-1,1) --++(1,1) -- ++(-1,1) --++(1,1) -- ++(-1,1) --++(1,1) -- ++(6,-6);
    \node at (3,5) {$\lambda$};
    \begin{scope}[xshift=-0.5em,yshift=-0.6em]
    \node[blue] at (0.5,1) {$5$};
    \node[blue] at (1.5,2) {$4$};
    \node[blue] at (2.5,3) {$\bar1$};
    \node[blue] at (3.5,4) {$\bar2$};
    \node[blue] at (7.5,2) {$\bar3$};
    \end{scope}
    \draw[dashed,blue,<->](0.5,0.5) -- (9.5,0.5);
    \draw[dashed,blue,<->](1.5,1.5) -- (6.5,1.5);
    \draw[dashed,blue,<->](2.5,2.5) -- (5.5,2.5);
    \draw[dashed,blue,<->](3.5,3.5) -- (4.5,3.5);
    \draw[dashed,blue,<->](7.5,1.5) -- (8.5,1.5);
    \begin{scope}[xshift=0.5em,yshift=-0.6em]
    \node[red] at (9.5,1) {$5$};
    \node[red] at (6.5,2) {$4$};
    \node[red] at (5.5,3) {$\bar1$};
    \node[red] at (4.5,4) {$\bar2$};
    \node[red] at (8.5,2) {$\bar3$};
    \end{scope}
    \end{tikzpicture}
    \caption{Reading off $w^\lambda=\bar2\bar14\bar35$ for $\lambda=(6,2,1)$ and $(d,r)=(3,2)$}
    \label{fig:read-wlambda}
\end{figure}
\end{remark}
\end{comment}

\subsection{Bijection between \texorpdfstring{$\BS(\lambda)$}{} and \texorpdfstring{$\BS(Z(d,r))|_{\lambda}$}{}}

Recall some notations from the beginning of this section: $\mu_i=Z(d,r)_i-\lambda_i$, $\sigma_i=\sum_{k=1}^i\mu_k$ and $N=|\lambda|$. 
\begin{defin}
    Define $\BS(Z(d,r))|_{\lambda}$ to be the set of balanced tableaux $T$ of shape $Z(d,r)$ such that for all $i\in [d]$ and any $k\in [N+\sigma_{i-1}+1,N+\sigma_i]$, $k$ appears in row $i$ of $T$.
\end{defin}

\begin{lemma}\label{lemma:swap-add}
    Let $B\in \BS(\lambda)$ and fix some $i\in [d]$ such that either $i=1$ or $\lambda_{i-1}\geq\lambda_i+3$. Denote $\lambda^{\#}$ the shifted diagram obtained from $\lambda$ by adding a box in the $i$-th row. Let $j$ be the column index of the box $\lambda^{\#}\setminus\lambda$.
    Let $B^{\#}$ be the tableau obtained from $B$ by 
    \begin{enumerate}
        \item interchange column $j$ and $j+1$ of $B$,
        \item define $B^{\#}(i,j) = N+1$.
    \end{enumerate}
    Then $B^{\#}$ is a balanced tableau and the following map is a bijection:
    \begin{align}\label{eqn:swap&add}
    \begin{split}
        f_i:\BS(\lambda) \longrightarrow &\{T\in \BS(\lambda^{\#}):T(i,j) = N+1\}\\
        B \longmapsto &B^{\#}.
    \end{split}
    \end{align}
\end{lemma}

\begin{lemma}\label{lm:BStoBSlambda}
    $\BS(Z(d,r))|_{\lambda}$ is the image of $\BS(\lambda)$ under the composition of maps $F= (f_d)^{a_d}\circ (f_{d-1})^{a_{d-1}}\circ\cdots\circ (f_1)^{a_1}$ with each $f_i$ defined as in \eqref{eqn:swap&add}. As a result, $F$ is a bijection between $\BS(\lambda)$ and $\BS(Z(d,r))|_{\lambda}$.
\end{lemma}

\subsection{Bijection between \texorpdfstring{$\BS(Z(d,r))|_{\lambda}$}{} and \texorpdfstring{$\Red(w^{(d,r)})|_{\lambda}$}{} }

\begin{prop}\label{lm:BSlambdatoRed}
    Let $\mathbf{a}\in \Red(w^{(d,r)})$ be a reduced word. Then $\ro(\mathbf{a})$ gives a balanced tableau in $\BS(Z(d,r))|_{\lambda}$ if and only if the ending segment of $\mathbf{a}$ is the same as $\mathbf{a}^\lambda$ as in \cref{def:alambda}. Consequently, this induces a bijection between $\BS(Z(d,r))|_{\lambda}$ and $\Red(w^{(d,r)})|_{\lambda}$.
\end{prop}

Combining \cref{prop:alambda}, \cref{lm:BStoBSlambda} and \cref{lm:BSlambdatoRed}, we get a bijection between $\SYT(\lambda)$ and $\BS(\lambda)$.

\begin{prop}
The bijection $\SYT(\lambda)\longleftrightarrow\BS(\lambda)$ in \cref{thm:main} does not depend on the parameter $r$.
\end{prop}

\begin{ex}\label{ex:main-ex}
We now work out an example in the case where $\lambda=(6,2,1)$, $d=3$ and $r=2$. Assume we start with a balanced tableau $B\in\BS(\lambda)$ shown here: \[\ytableausetup{boxsize=1.4em}
B = \begin{ytableau}
6 & 3 & 4 & 1 & 5 & 9 \\
\none & 7 & 8 & \none & \none \\
\none & \none & 2 & \none 
\end{ytableau}\, .\]
We can complete it to $B^+\in \BS(Z(3,2))$ using the algorithm in \cref{lm:BStoBSlambda}
\begin{center}
\ytableausetup{boxsize=1.5em}
$B = $
\begin{ytableau}
    6 & 3 & 4 & 1 & 5 & 9 \\
    \none & 7 & 8 & \none & \none \\
    \none & \none & 2 & \none
\end{ytableau}
$\rightarrow$ 
\begin{ytableau}
    6 & 3 & 4 & 1 & 5 & 9 & \circled{10} \\
    \none & 7 & 8 & \none & \none \\
    \none & \none & 2 & \none
\end{ytableau}
$\rightarrow$
\begin{ytableau}
    6 & 3 & 4 & 5 & 1 & 9 & \circled{10} \\
    \none & 7 & 8 & \circled{11} & \none \\
    \none & \none & 2 & \none
\end{ytableau}
\vspace{0.2cm}
$\rightarrow$
\begin{ytableau}
    6 & 3 & 4 & 5 & 9 & 1 & \circled{10} \\
    \none & 7 & 8 & \circled{11} & \circled{12} \\
    \none & \none & 2 & \none
\end{ytableau}
$\rightarrow$
\begin{ytableau}
    6 & 3 & 4 & 5 & 9 & \circled{10} & 1 \\
    \none & 7 & 8 & \circled{11} & \circled{12} & \circled{13}\\
    \none & \none & 2 & \none
\end{ytableau}
$\rightarrow$
\begin{ytableau}
    6 & 3 & 4 & 9 & 5 & \circled{10} & 1 \\
    \none & 7 & 8 & \circled{12} & \circled{11} & \circled{13}\\
    \none & \none & 2 & \circled{14}
\end{ytableau}
\vspace{0.2cm}
$\rightarrow$
\begin{ytableau}
    6 & 3 & 4 & 9 & \circled{10} & 5 & 1 \\
    \none & 7 & 8 & \circled{12} & \circled{13} & \circled{11}\\
    \none & \none & 2 & \circled{14} & \circled{15}
\end{ytableau}
$= B^+$
\end{center}
Now $B^+$ gives a reflection order of $w^{(d,r)}$ as follows
\[
\begin{tikzcd}[row sep = small]
12345 \arrow[r, "e_3-e_2"]& 13245 \arrow[r, "e_1"] & \bar13245 \arrow[r, "e_3+e_1"] & 3\bar1245 \arrow[r, "e_3"] & \bar3\bar1245 \arrow[r, "e_3-e_1"] & \bar1\bar3245\\
\arrow[r, "e_3+e_2"]& \bar12\bar345 \arrow[r, "e_2+e_1"] & 2\bar1\bar345 \arrow[r, "e_2"] & \bar2\bar1\bar345 \arrow[r, "e_4+e_3"] & \bar2\bar14\bar35 \arrow[r, "e_5+e_3"] & \bar2\bar145\bar3\\
\arrow[r, "e_2-e_1"]& \bar1\bar245\bar3 \arrow[r, "e_4+e_2"] & \bar14\bar25\bar3 \arrow[r, "e_5+e_2"] & \bar145\bar2\bar3 \arrow[r, "e_4+e_1"] & 4\bar15\bar2\bar3 \arrow[r, "e_5+e_1"] & 45\bar1\bar2\bar3.\\
\end{tikzcd}
\]
We can read off the reduced word $\mathbf{a}=201012103412312\in\Red(w^{(3,2)})$. We can confirm that the reduced word ends with $\mathbf{a}^\lambda=412312$. Now we perform the Kra\'skiewicz's insertion on $\mathbf{a}$ described in \cref{sec:kraskiewicz} and we get
\[\ytableausetup{boxsize=1.4em}
P(\mathbf{a}) = \begin{ytableau}
4 & 3 & 0 & 1 & 2 & 3 & 4\\
\none & 3 & 0 & 1 & 2 & 3\\
\none & \none &0 & 1 & 2
\end{ytableau},\ 
Q(\mathbf{a}) = \begin{ytableau}
1 & 2 & 3 & 5 & 6 & 9 & 10\\
\none & 4 & 7 & 11 & 12 & 13\\
\none & \none & 8 & 14 & 15
\end{ytableau},
\]
Finally, let $T^+=Q(\mathbf{a})\in\SYT(w^{(3,2)})$, and $T\in\SYT(\lambda)$ is obtained from $T^+$ by deleting the largest entries until $|\lambda|$ entries are left:
\[\ytableausetup{boxsize=1.5em}
T^+ = \begin{ytableau}
1 & 2 & 3 & 5 & 6 & 9 & \circled{10}\\
\none & 4 & 7 & \circled{11} & \circled{12} & \circled{13}\\
\none & \none & 8 & \circled{14} & \circled{15}
\end{ytableau},\ 
T = \begin{ytableau}
1 & 2 & 3 & 5 & 6 & 9\\
\none & 4 & 7\\
\none & \none & 8
\end{ytableau}.
\]
\end{ex}

.