Justify the following running time

Question

The book "Cracking the Coding Interview , 6th ed." describes the following method to sum the values of all nodes in a balanced binary search tree, and also claims that this method runs in O(n) time.

int sum(Node node) {
    if (node == null) {
        return 0;
    }
    return sum(node.left) + node.value + sum(node.right);
}

It seems to me that this method iterates through every node in the tree exactly once, plus one null reference for each leaf in the tree. I would say that n is the number of nodes in the tree, and we visit O(n+b) nodes (where b is the number of leaf nodes, is the number of null nodes traversed) and do constant work there.

How can n be an upper bound if we visit (or attempt to visit) more than n nodes? You might argue that b is constant and so we drop it, but b grows proportionally to n.

Please don't mark this as a duplicate just to get your internet points. Specific questions have specific answers.

Answer 1

The book is correct, because $b\leq 2n$, so you visit no more than $3n$ nodes, which is $O(n)$, and $b\leq 2n$ because no node has more than two leaves. You correctly say that $b$ grows proportional to $n$, which is exactly the key.

Just for clarity, Big-Oh notation means "grows proportional to, ignoring constant factors" (roughly), which is why we say that $3n\in O(n)$, even though one is a multiple of the other, and $6n^2+3n+42\in O(n^2)$ and $42e^n\in O(e^n)$, for example.

(actually, $O(n)$ means, "is not more than $cn$". For "is proportional to $n$", use $\Theta(n)$ instead of $O(n)$).