A Red-Black tree is a self-balancing binary search tree. The main difference to
a normal binary tree is an extra bit that stores the parity (the *color*) of
each node. This bit can be either be 0 (*black*) or 1 (*red*). An example of a
red-black tree is shown in Figure 1 (Hardcoded)

The following invariant must be satisfied every time an operation is being made on node of a Red-Black tree:

- Every node is either red or black
- The root is always black
- Every leaf is black
- If a node is red, then both its children are black
- All paths from a node to descendant nodes contain the same number of black nodes

The node of a Red-Black tree can be described by the following data type:

In [ ]:

```
from enum import Enum
class RBNode:
def __init__(self, color, key, left, right, parent):
self.left = left
self.right = right
self.parent = parent
self.key = key
self.color = color
def __str__(self):
return str("(%s)%s" % (self.color, self.key))
class NodeColors(Enum):
BLACK = 0;
RED = 1;
```

The data structure of the tree is nothing more than a collection of nodes with the root node as the starting point. The constructor initializes a root node to a provided value

In [ ]:

```
class RBTree:
def __init__(self, key):
self.nil = RBNode(NodeColors.BLACK, None, None, None, None)
self.root = RBNode(NodeColors.BLACK, key, self.nil, self.nil, self.nil)
```

Your task for this assignment is to construct algorithms that manipulate Red-Black trees.

We start constructing a Red-Black tree by calling the constructor

`tree = RBTree(5)`

Inserting a node consists of three steps:

- Finding the appropriate branch to insert the node in the tree
- Inserting the node and creating links to the tree
- Rebalancing the tree

For Step 1, the procedure is not much different from a normal binary search
tree. Specifically, the algorithm first finds the appropriate position
for the new node by searching through the tree. In the following snippet,
`z`

is the value we would like to insert. However, the snippet has a
mistake.

**T1 (10 points)**: Spot and fix the mistake in the `search`

method

In [ ]:

```
def search(tree, z):
y = tree.nil
x = tree.root
while x != tree.nil:
y = x
if z.key < x.key:
x = x.left
else:
y = x.right
return y
```

Using the (hopefully correct by now!) `search`

method, we can locate the node
to insert our new node as a child. This happens by creating the node and
linking it to the appropriate super/sub nodes. Note that for new nodes,
the starting color is always `RED`

.

**T2 (20 points)**: Write code to construct and link a new node.
*Hint*: Use the `search`

method inside the `insert`

method.
*Hint*: Make the method return the new node, this is usefull for later

In [ ]:

```
def insert(tree, z):
pass
```

In [ ]:

```
#Test you implementation here
```

After inserting the node, the tree potentially needs to be rebalanced. An example of why this is so can be seen in the following plot. What we see here is that the insertion of node 4 violates invariant number 4.

The rebalancing operation can be quite complicated; the pseudocode below takes care of all the cases.

```
proceduce rb_balance(T, z)
'' T is an RBTree
'' z is the newly inserted node
while z.parent.color == RED
if z.parent == z.parent.parent.left
y <- z.parent.parent.right
if y.color == RED
z.parent.color <- BLACK
y.color <- BLACK
z.parent.parent.color <- RED
z <- z.parent.parent
else
if z == z.parent.right
z <- z.parent
left_rotate(T, z)
z.parent.color <- BLACK
z.parent.parent.color <- RED
right_rotate(T, z.parent.parent)
else
y <- z.parent.parent.left
if y.color == RED
z.parent.color <- BLACK
y.color <- BLACK
z.parent.parent.color <- RED
z <- z.parent.parent
else
if z == z.parent.left
z <- z.parent
right_rotate(T, z)
z.parent.color <- BLACK
z.parent.parent.color <- RED
left_rotate(T, z.parent.parent)
T.root.color <- BLACK
```

You might be wondering with `left_rotate`

and `right_rotate`

are. They are
functions that represent common operations on trees. To understand how they work
have a look at the plot below.

Below you can find the Python code for the left_rotate method.

*Note*: The node you give to the rotate methods is always the parent node of the 2 nodes that are rotating

In [ ]:

```
def left_rotate(tree, x):
y = x.right
x.right = y.left
if y.left != tree.nil:
y.left.parent = x
y.parent = x.parent
if x.parent == tree.nil:
tree.root = y
elif x == x.parent.left:
x.parent.left = y
else:
x.parent.right = y
y.left = x
x.parent = y
```

**T3 (20 points)**: Implement the code for `right_rotate`

In [ ]:

```
def right_rotate(tree, x):
pass
```

In [ ]:

```
#Test your implementation here
```

**T4 (25 points)**: Implement the rebalancing pseudocode in Python

In [ ]:

```
def balance(tree, z):
pass
```

In [ ]:

```
#Test you implemention here
```

**T5 (25 points)**: Create a method that will
exhaustively search whether the five invariants described above are maintained in the tree.

In [ ]:

```
def check():
"@return: True, only if tree satisfies all the Red-Black tree invariants"
pass
```

In [ ]:

```
#Test all of the five invariants here
```

**T6 (10 points)**: Prove that the average and worse case complexity for the
insertion in Red-Black trees is $O(log(n))$.

- A Red-Black tree in Java
- A Red-Black tree in C
- Wikipedia on Red-Black Trees