The remaining target 4 is sufficient for an existence question with these fixed rules. If you must output the path itself, you also need a way to reconstruct its nodes.

Let has_sum(node, remaining) ask whether a path beginning at node and ending at a leaf sums to remaining. The current node has not yet been charged.

def has_sum(node, remaining):
    if node is None:
        return False
    value, left, right = node
    remaining -= value
    if left is None and right is None:
        return remaining == 0
    return has_sum(left, remaining) or has_sum(right, remaining)

tree = (8, (3, None, None), (10, None, None))
assert has_sum(tree, 11)
assert not has_sum(tree, 8)
assert not has_sum(None, 0)

Why is reaching zero at the root not enough? The contract requires a leaf, not merely any prefix. Why is None, 0 false? In this problem, an empty tree has no root-to-leaf path. Base cases express the requested output semantics.

Worst-case time is O(n) and stack space O(h). Short-circuiting can stop early when one branch succeeds.

This information-flow question generalizes: validating a BST passes allowed bounds from ancestors; computing height returns child summaries. Some problems need both directions. Choose the contract before choosing preorder or postorder terminology.

Analogy limit: passing only a remaining sum works because past values affect the answer only through their sum. A rule such as “no repeated value on the path” needs additional context.

No. A later negative value can make the required sum possible. That pruning rule requires appropriate restrictions, such as all remaining node values being nonnegative.