# Problem 13.2

## Join operation on red-black trees

The

joinoperation takes two dynamic sets $S_1$ and $S_2$ and an element $x$ such that any $x_1 \in S_1$ and $x_2 \in S_2$, we have $x_1.key \le x.key \le x_2.key$. It returns a set $S = S_1 \cup \{x\} \cup S_2$. In this problem, we investigate how to implement the join operation on red-black trees.

- Given a red-black tree $T$, let us store its black-height as the new attribute $T.bh$. Argue that
`RB-INSERT`

and`RB-DELETE`

can maintain the $bh$ attribute without requiring extra storage in the nodes of the tree and without increasing the asymptotic running times. Show that while descending through $T$, we can determine the black-heigh of each node we visit in $\O(1)$ time per node visited.We wish to implement the operation $\text{RB-JOIN}(T_1, x, T_2)$, which destroys $T_1$ and $T_2$ and returns a red-black tree $T = T_1 \cup \{x\} \cup T_2$. Let $n$ be the total number of nodes in $T_1$ and $T_2$.

- Assume that $T_1.bh \ge T_2.bh$. Describe an $\O(\lg n)$-time algorithm that finds a black node $y$ in $T_1$ with the largest key from among those nodes whose black height is $T_2.bh$.
- Let $T_y$ be the subtree rooted at $y$. Describe how $T_y \cup \\{x\\} T_2$ can replace $T_y$ in $\O(1)$ time without destroying the binary search tree property.
- What color should we make $x$ so that red-black properties 1, 3, and 5 are maintained? Describe how to enforce properties 2 and 4 in $\O(\lg n)$ time.
- Argue that no generality is lost by making the assumption in part (b). Describe the symmetric situation that arises when $T_1.bh \le T_2.bh$.
- Argue that the running time of
`RB-JOIN`

is $\O(\lg n)$.

### a. Maintaining black height in constant time

Technically speaking, whenever we perform the fixup operations, we can keep track of the colors we change in the path we've modified, and rely on that to know whether we have to increment or decrement the black height.

But it seems to me, that there is a simpler heuristic, that I'm going to just guess here, without proving. Specifically, whenever the black height changes, it is somehow reflected in the second and third layer of the tree.

When inserting, the way the black height increases is by coloring the root red
(while preserving the properties with `RB-INSERT-FIXUP`

) and then finally
coloring it black. This is the only operation that increases the black height by
one.

Similarly, when deleting, the extra-black pointer eventually finds itself to the root, and that's how the black heigh gets decreased.

I may be missing some cases here, but fundamentally - both left and right subtrees of the root need to lose or gain a unit of black height in order for the tree to do so as well, hence we can figure it out by observing what happens around the root.

### b. Finding a black node with the largest key with a specific black-height

Quite simply, we start from the root and go right, setting $c = T.bh$. Every time we go through a black node, we decrement $c$ until we reach the $T_2.bh$. Once we do, we have the node in question.

### c. Replacing $T_y$

Quite simply, we replace the node with $x$, and we put $T_y$ as its left child, and $T_2$ as it's right child. The binary search tree invariant is preserved, as we have that $x.key$ is smaller than the keys in $T_2$ and larger than the ones in $T_1$.

### d. What color should we make $x$?

We should make it red. This preserves properties 1, 3 and 5.

Two problems can occur now. Either it's the new root, and it's red, in which case we can color it black and be done with it, or it can have a red parent (its left child is black, because we found a black node for $T_y$ and its right child is also black, because it's $T_2$, and the root of $T_2$ is black).

We then simply need to call `RB-INSERT-FIXUP`

to fix the two subsequent red
nodes.

### e. Generality

Pretty trivially, if $T_2$ has the greater black height, we need to find the node with the smallest key of a given black height, replace it with $x$ and put $T_1$ as it's left child. The approach is symmetrical.

### f. Running time

The operation does the following:

- Descends one of the trees to find $T_y$, and does so in $\O(\lg n)$ time.
- Replaces it with $x$ and transplants $T_y$ and $T_2$ under it, in $\O(1)$ time.
- Runs
`RB-INSERT-FIXUP`

, which takes $\O(\lg n)$ time.

We get $\O(\lg n) + \O(1) + \O(\lg n) = \O(\lg n)$.