De Casteljau's algorithm without temporaries / stack
The OP has indicated in the comments that they are only interested in a specific time $t$. De Casteljau's algorithm can be described as follows:


Input: order $N$, points $A_0,\ldots,A_N$, time $t \in (0,1)$.
Calculate $\overline{A_i A_{i+1}} = A_i(1-t) + A_{i+1}t$ for $0 \leq i \leq N-1$.
Calculate $\overline{A_i A_{i+1} A_{i+2}} = \overline{A_i A_{i+1}} (1-t) + \overline{A_{i+1} A_{i+2}} t$ for $0 \leq i \leq N-2$.
Continue similarly for $N-2$ more steps.
Output $\overline{A_0 A_1 \ldots A_n}$.


It is not completely clear what it does it meant to "implement" De Casteljau's algorithm. However, a reasonable informal interpretation of this statement is as follows:


  Schedule the computations in De Casteljau's algorithm in a way that uses the least amount of space and the least amount of computations.


Note that the two efficiency measures could potentially conflict.

There is an area of theoretical computer science covering exactly such questions, namely pebbling. See Jakob Nordström's survey-in-making. Your task exactly corresponds to pebbling a height $N$ pyramid. On the one hand, this is known to require $N+2$ pebbles (roughly temporary memory locations). On the other hand, there is a pebbling strategy using $N+2$ pebbles and a linear number of steps (a linear number of computations &mdash; we are allowing computing each node, i.e. $\overline{A_i \ldots A_j}$, more than once).

Here is how this pebbling strategy looks like in the case $N=2$:


$T_1 \gets A_0$.
$T_2 \gets A_1$.
$T_3 \gets T_1(1-t) + T_2t$.
$T_1 \gets A_2$.
$T_4 \gets T_2(1-t) + T_0t$.
$T_1 \gets T_3(1-t) + T_4t$.


The output is in $T_1$.