Iterating Over Binary Trees

2012-04-19 Tavian Barnes

Binary trees are great. If you ever have to implement one yourself though, you're probably either using C or you need to look at the documentation for your language's standard library more closely. Even POSIX C has the tsearch family of functions from <search.h>.

Recently I found myself attempting to implement a bootstrapping compiler in a rather restrictive subset of Java, and found the need for some kind of associative array data structure. I chose a splay tree, because, well, I like them.

Soon I found that I wanted the ability to iterate over the entire tree, in sorted order. This ability is provided by C++'s std::map::iterators, for example. I considered using a threaded tree, then an iterator holding a stack, before I realised that there is a much simpler solution with equal performance. It requires only that the nodes store a reference to their parent node, which mine could with little extra effort.

The principle is simple. If a node has a right subtree, its successor is the minimal (leftmost) node in that subtree. Otherwise, its successor is the parent of its first ancestor which is not a right child. In code,

public class Node {
    private Node left, right;

    public Node next() {
        Node next;

        if (right == null) {
            // Node has no right child
            next = this;
            while (next.parent != null && next == next.parent.right) {
                next = next.parent
            next = next.parent;
        } else {
            // Find the leftmost node in the right subtree
            next = right;
            while (next.left != null) {
                next = next.left;

        return next;

But how fast is this? An individual call to could traverse from the bottom to the top of the tree, at a cost of O(h)O(h), where hh is the height of the tree. If there are nn nodes in the tree, could we traverse O(nh)O(n h) edges to visit the whole tree? This would be particularly bad for a splay tree, which is not necessarily balanced at any given time.

It turns out that we need only traverse O(n)O(n) edges to visit the entire tree. For an informal proof, note that our algorithm is basically equivalent to tracing an outline of the tree like this:

Clearly we can only trace around any edge at most twice (on its left and right side). Since any tree has n1n - 1 edges, we traverse at most 2(n1)O(n)2 (n - 1) \in O(n) edges in the process of exploring the whole tree.