= Solution
Use `r0` for the argument and result. With 32-bit <integers> and <pointers>, the three fields have byte offsets 0, 4 and 8. Assume an acyclic finite <binary tree>, no concurrent mutation, a valid aligned <call stack>, and no signed <integer overflow> in any evaluated sum. These are representation and arithmetic assumptions, rather than properties of arbitrary <C programming language> implementations.
``
.syntax unified
.cpu cortex-m0
.thumb
.global sum_rec
.thumb_func
sum_rec:
cmp r0, #0
beq .Lempty
push {r4, r5, r6, lr} @ 16-byte frame; r6 is alignment padding
movs r4, r0 @ retain this invocation's node
ldr r5, [r4, #0] @ running value begins with its label
ldr r0, [r4, #4]
bl sum_rec @ r0 = left subtree sum
adds r5, r5, r0
ldr r0, [r4, #8]
bl sum_rec @ r0 = right subtree sum
adds r0, r0, r5
pop {r4, r5, r6, pc} @ restore caller state and return
.Lempty:
movs r0, #0
bx lr
``
The node and partial sum are held in <callee-saved registers> `r4,r5`. Each recursive invocation saves its caller's values before replacing them, so a child returns with the parent's values intact. Saving `lr` preserves the return address across the two `bl` instructions. <Caller-saved registers> may be overwritten by a child; no parent value is left live only in them. Pushing four words maintains eight-byte <stack alignment> at recursive calls.<Procedure Call Standard for the Arm Architecture by ARM Limited>[AAPCS32, base standard: core-register roles, stack constraints and result return]{ref}
Correctness follows by <structural induction>. An empty subtree returns zero. For a nonempty subtree, assume both recursive calls return their respective label sums. The final addition then combines the node's label with exactly those two sums, so \b[`r0` returns the sum of all labels]. The instruction forms and field offsets fit the short load and arithmetic encodings.<Thumb 16-bit instruction set quick reference by Arm>[Move, Add, Load, Push/Pop and Branch tables]{ref}
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