Lowest Common Ancestor of a Binary Tree with parent pointer

Given a binary tree, find the lowest common ancestor of two given nodes in the tree. Each node contains a parent pointer which links to its parent.

Note:
This is Part II of Lowest Common Ancestor of a Binary Tree. If you need to find the lowest common ancestor without parent pointers, please read Lowest Common Ancestor of a Binary Tree Part I.

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

 An easy solution:   use HashSet

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.

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
}