The Union Operation

To union nodes $a$ and $b$:

$1.$ Find the leader of $a$. Call it rootA.

$2.$ Find the leader of $b$. Call it rootB.

$3.$ If rootA == rootB, they are already in the same group. Done.

$4.$ Otherwise, set parent[rootA] = rootB (or vice versa).

Now everyone in $a$'s old group points (eventually) to $b$'s leader. The two groups are merged. This takes $O(h)$ for the two find calls plus $O(1)$ for the pointer update.

Space complexity is $O(n)$ for the data structures used.