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To add two finite floating-point numbers, compare their exponents, shift the smaller significand to a common exponent, add or subtract the signed significands, normalize the result, and apply floating-point rounding. Extra low-order bits are needed to determine correct rounding when shifting or cancellation discards information. A carry beyond the leading digit increases the exponent; cancellation can decrease it. Unlike exact real addition, this operation need not be associative. Changing the child evaluation order in a fixed binary expression tree need not change its parenthesization, whereas reassociation changes which intermediate results undergo floating-point rounding.

Suppose each subtree of a binary expression tree is evaluated completely before its sibling, leaves one result on an operand stack, and is not spilled or interleaved. If the child depths are $p,q$, evaluating the first child needs $p$ slots and keeping its result while evaluating the second needs $1+q$. Reversing the order gives $\max (q,1+p)$. If either order preserves the expression's meaning, the minimum is $\max (p,q)$ when $p\ne q$ and $p+1$ when $p=q$. For example, if $p>q$, integrality gives $p\ge q+1$, so evaluating the deeper child first attains the lower bound $p$. If $p=q$, the second child needs its own $p$ slots plus one saved result regardless of order. Induction over the binary expression tree proves optimality in this evaluation model.

For a noncommutative operator, an instruction swapping the top two stack entries or a reverse-operand opcode permits the same ordering while preserving operand positions. Reordering still requires absence of order-sensitive side effects. Reassociation and interleaving are separate transformations and are not part of this optimality claim.

Let $d_e=\mathrm{depth}(e)$ and $d_f=\mathrm{depth}(f)$ measure the extra operand stack space required from the current base, with each evaluation leaving one result. Evaluating $e$ first uses $d_e$ slots. Its result occupies one slot while evaluating $f$, so that stage uses $1+d_f$ slots. The final add reduces two slots to one. ThusBy commutativity of addition, choose the smaller of these two bounds. Evaluate the child of larger depth first. The optimum recurrence on a binary expression tree isAssign depth one to a literal or a simple already-addressable value, and compute the child depths recursively. If $d_e>d_f$, the depths are integers, so $d_e\ge 1+d_f$; evaluating $e$ first therefore attains $d_e$, while reversing the order needs $d_e+1$. If the depths are equal, whichever child goes second needs the extra retained-result slot. These lower bounds prove the recurrence is optimal for evaluating each child subtree without interleaving or spilling. Apply it at every node to minimize peak operand stack occupancy for the given binary expression tree; this is optimal stack evaluation of a binary expression tree.

This changes child evaluation order, not the parenthesization. For example, a chain-shaped binary expression tree can have depth two, whereas a balanced tree of eight leaves has depth four under this model. The reordering assumes expressions have no order-sensitive side effects or observable exceptions. Reassociating floating-point addition would require additional care because it can change floating-point rounding.