Question

I know similar questions have been asked before, but I think my solution is far simpler. Especially compared to Wikipedia.

Please prove me wrong!

If you have a tree with nodes that have the given data structure:

struct node
{
    node * left;
    node * right;
    node * parent;
    int key;
}

You could write a function like this:

node* LCA(node* m, node* n)
{
    // determine which of the nodes is the leftmost
    node* left = null;
    node* right = null;
    if (m->key < n->key)
    {
        left = m;
        right = n;
    }
    else
    {
        left = n;
        right = m;
    }
    // start at the leftmost of the two nodes,
    // keep moving up the tree until the parent is greater than the right key
    while (left->parent && left->parent->key < right->key)
    {
        left = left->parent;
    }
    return left;
}

This code is pretty straightforward and worst case is O(n), average case it is probably O(logn), especially if the tree is balanced (where n is the number of nodes in the tree).

Answer 1

Your Algorithm looks okay to me, at least I couldn’t think of anything better. Note that you don’t need the parent pointer; instead you can go down the tree starting from the root, and find the first node whose key lays between the two initial keys.

However, your problem has nothing to do with the one Tarjan solved. First of all, you consider binary trees and he considers n-ary trees; but this is probably a detail. More importantly, you consider search trees, while Tarjan considers general trees (no ordering on the keys). Your problem is much simpler, because, depending on the key, you can guess where a certain node has to be in the tree.