Lowest Common Ancestor of a Binary Tree "WITH" Parent pointers

[sourcecode]
          ___3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

Using the tree above as an example, the LCA of nodes 5 and 1 is 3. Please note that LCA for nodes 5 and 4 is 5.

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An easy solution:
As we trace the two paths from both nodes up to the root, eventually it will merge into one single path. The LCA is the exact first intersection node where both paths merged into a single path. An easy solution is to use a hash table which records visited nodes as we trace both paths up to the root. Once we reached the first node which is already marked as visited, we immediately return that node as the LCA.

[sourcecode]
Node *LCA(Node *root, Node *p, Node *q)
{
    hash_set <Node *> visited;
    while (p || q)
    {
        if (p)
        {
            if (!visited.insert(p).second)
            return p; // insert p failed (p exists in the table)
            p = p->parent;
        }
        if (q)
        {
            if (!visited.insert(p).second)
            return q; // insert q failed (q exists in the table)
            q = q->parent;
        }
    }
    return NULL;
}
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The run time complexity of this approach is O(h), where h is the tree’s height. The space complexity is also O(h), since it can mark at most 2h nodes.

The best solution:
A little creativity is needed here. Since we have the parent pointer, we could easily get the distance (height) of both nodes from the root. Once we knew both heights, we could subtract from one another and get the height’s difference (dh). If you observe carefully from the previous solution, the node which is closer to the root is always dh steps ahead of the deeper node. We could eliminate the need of marking visited nodes altogether. Why?

The reason is simple, if we advance the deeper node dh steps above, both nodes would be at the same depth. Then, we advance both nodes one level at a time. They would then eventually intersect at one node, which is the LCA of both nodes. If not, one of the node would eventually reach NULL (root’s parent), which we conclude that both nodes are not in the same tree. However, that part of code shouldn’t be reached, since the problem statement assumed that both nodes are in the same tree.

[sourcecode]
int getHeight(Node *p) 
{
    int height = 0;
    while (p) 
    {
        height++;
        p = p->parent;
    }
    return height;
}

// As root->parent is NULL, we don't need to pass root in.
Node *LCA(Node *p, Node *q) 
{
    int h1 = getHeight(p);
    int h2 = getHeight(q);
    // swap both nodes in case p is deeper than q.
    if (h1 > h2) 
    {
        swap(h1, h2);
        swap(p, q);
    }
    // invariant: h1 <= h2.
    int dh = h2 - h1;
    for (int h = 0; h < dh; h++)
    q = q->parent;
    while (p && q) 
    {
        if (p == q) return p;
        p = p->parent;
        q = q->parent;
    }
    return NULL; // p and q are not in the same tree
}
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