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Incoming links: Explicit-stack depth-first tree traversal
Explicit-stack depth-first tree traversal 2026-10-24
An explicit-stack depth-first tree traversal stores pending subtree roots instead of suspended recursive calls. In a binary tree, visit the current node, defer its nonempty right child on a stack, and continue with its left child; pop when the current subtree is empty. The unvisited subtrees remain disjoint and, together with the visited nodes, partition the original tree. This loop invariant proves that accumulating each visited label gives the tree sum. At most one right sibling per ancestor is pending, so auxiliary storage is for depth and running time is for nodes. Unlike a full recursive stack frame, a pending entry need contain only a subtree pointer.
Push the right child before the left child so that the LIFO stack exposes the left child first:Let be the sequence of labels in the preorder traversal of . If the abstract pending stack, top first, is , the loop invariant is that the output already printed, followed by , is the required full output. Each pending root is non-null, and its subtree is still unvisited. This is the pending-work form of explicit-stack depth-first tree traversal.
def preorderIterative(t: Tree): Unit = {
val pending = new Stack
if (t != null) pending.push(t)
while (!pending.isEmpty) {
val u = pending.pop
println(u.datum)
if (u.right != null) pending.push(u.right)
if (u.left != null) pending.push(u.left)
}
}One iteration discharges the first root's own label and replaces its remaining obligation by the left subtree followed by the right subtree. The push order makes that precisely the new top-first pending order. Thus the loop invariant is preserved. The loop variant is the sum of the numbers of vertices in all pending subtrees; it decreases by exactly one per iteration, even if the number of stack entries increases. When the stack is empty there is no remaining output obligation, proving correctness and termination. This assumes a finite genuine binary tree and, as permitted, sufficient stack capacity. Each vertex is pushed and popped once, giving time and pending references for binary tree height .
Use an explicit-stack depth-first tree traversal. Descend the left branch immediately and put each nonempty right subtree on a stack for later. For the same node declarations as in the paper, an original C programming language implementation is:
int sum_iter(tree t)
{
tree pending[32];
unsigned top = 0;
int total = 0;
while (t != NULL || top != 0) {
if (t == NULL) {
t = pending[--top];
} else {
total += t->data;
if (t->right != NULL)
pending[top++] = t->right;
t = t->left;
}
}
return total;
}The loop invariant says that the accumulated labels plus the sums of the current subtree and all pending subtrees equal the original tree sum, with the unvisited subtrees disjoint. Visiting a node adds its label and replaces its subtree by its two children. Popping merely changes which pending subtree is current. When both current and pending subtrees are absent, the invariant proves the result.
At most one unvisited right sibling is pending at each level of the current ancestor path. If depth means the number of nonempty nodes on a longest root-to-leaf path, depth at most 32 requires at most 31 pending right siblings. A 32-word array is therefore sufficient; it also covers the convention that depth counts at most 32 edges. The depth promise makes unchecked pushes valid here. For arbitrary input, a production implementation should check capacity rather than overwrite its frame.
A corresponding Thumb instruction set implementation keeps the total in
r4, the array base in r5, and its next-free address in r6:.global sum_iter_asm
.thumb_func
sum_iter_asm:
push {r4, r5, r6, lr}
sub sp, sp, #128 @ 32 pending pointers
mov r5, sp @ base of pending array
mov r6, sp @ empty stack: next-free equals base
movs r4, #0
.Lvisit:
cmp r0, #0
beq .Ltake_pending
ldr r1, [r0, #0]
adds r4, r4, r1 @ consume this node's label
ldr r1, [r0, #8]
cmp r1, #0
beq .Lleft
str r1, [r6, #0] @ defer the nonempty right subtree
adds r6, r6, #4
.Lleft:
ldr r0, [r0, #4] @ continue down the left branch
b .Lvisit
.Ltake_pending:
cmp r6, r5
beq .Ldone
subs r6, r6, #4
ldr r0, [r6, #0]
b .Lvisit
.Ldone:
movs r0, r4
add sp, sp, #128
pop {r4, r5, r6, pc}The stack pointer stays fixed while the array is used;
r6 is a separate cursor inside that array. All callee-saved registers touched by the function are restored. Each nonempty node contributes its label exactly once. As in part (a), require that every partial total fits the signed result type; a representable final answer alone does not prevent overflow in a different summation order.