The problem being solved is that of how many registers does an N-ary node need in order to be evaluated.

First, we observe the numbers for the simplest case of a binary operator:

op1(n) OP(r) op2(m)

If n > m, we need n registers to evaluate "OP"
If n =< m, we need m + 1 registers, m for "op2" and one register to hold "op1" while "op2" is being evaluated

This is the same as saying r = max(n, m + 1)

The N-ary case can be reduced down to a series of binary ones, for example, for a 4-ary node we would have:

    OP
    /\
   /  \
  /    \
 op1   /\
      /  \
     /    \
   op2    /\
         /  \
        /    \
      op3    op4

Thus, to find the number of registers we would walk this tree in post-order, evaluating each binary node with the formula derived above.

Such a post-order walk is equivalent to the reversed linear loop over the operands implemented in this change.