Chapter 9

Chapter 9: Search Trees II

9.3 Multi-Way Search Trees

Definition

A node v of an ordered tree is a d-node if v has d children.
A multi-way search tree is an ordered tree such that:

Each internal node v has at least two children (d-node with d >= 2 )
v stores d − 1 key-element items (k_i, e_i) (d is the number of children)
For a node with children v₁, v₂, …, v_d storing keys k₁ < k_{2 <}… < k_d₋₁

keys in the subtree of v₁ are less than k₁
keys in the subtree of v_i are between k_i₋₁ and k_i (i = 2, …, d − 1)
keys in the subtree of v_d are greater than k_d₋₁

The leaves store no items and serve as placeholders

We can extend the notion of inorder traversal from binary trees to multi-way search trees
Namely, we visit item (k_i, e_i) of node v between the recursive traversals of the subtrees of v rooted at children v_i and v_i₊₁
An inorder traversal of a multi-way search tree visits the keys in increasing order

Searching in a Multi-Way Tree

Similar to search in a binary search tree
For each internal node with children v₁, v₂, …, v_d and keys k₁, k₂, …, k_d₋₁

k = k_i (i = 1, …, d − 1): the search terminates successfully
k < k₁: we continue the search in child v₁
k_i₋₁ < k < k_i (i = 2, …, d − 1): we continue the search in child v_i
k > k_d₋₁: we continue the search in child v_d

Reaching an external node terminates the search unsuccessfully
Example: search for 30

9.4 (2,4) Trees

A (2,4) tree (also called 2-4 tree or 2-3-4 tree) is a multi-way search tree with the following properties

Node-Size Property: every internal node has at most four children (and at least two)
Depth Property: all the external nodes have the same depth

Depending on the number of children, an internal node of a (2,4) tree is called a 2-node, 3-node or 4-node

Theorem: A (2,4) tree storing n items has height O(log n)
Proof:

Let h be the height of a (2,4) tree with n items
Since there are at least 2ⁱ items at depth i = 0, … , h −1 and no items at depth h, we have

n ≥ 1 + 2 + 4 +… + 2^h−1 = 2^h − 1
Thus, h ≤ log (n + 1)

Searching in a (2,4) tree with n items takes O(log n) time

9.4.1 Update Operations

Insertion

We insert a new item (k, e) at the parent v of the leaf reached by searching for k

We preserve the depth property but
We may cause an overflow (i.e., node v may become a 5-node)

Example: inserting key 30 causes an overflow

We handle an overflow at a 5-node v with a split operation:

let v₁, …, v₅ be the children of v and k₁, …, k₄ be the keys of v
node v is replaced nodes v' and v"
v' is a 3-node with keys k₁, k₂ and children v₁, v₂, v₃
v" is a 2-node with key k₄ and children v₄, v₅
key k₃ is inserted into the parent u of v (a new root may be created)

The overflow may propagate to the parent node u

Algorithm insertItem(k, e)
1.We search for key k to locate the insertion node v
2.We add the new item (k, e) at node v
3. while overflow(v)
        if isRoot(v)
            create a new empty root above v
        v ← split(v)

Analysis of Insertion in a (2,4) Tree

Let T be a (2,4) tree with n items

Tree T has O(log n) height
Searching takes O(log n) time because we visit O(log n) nodes
Restoring Node-Size Property takes O(log n) time because each split takes O(1) time and we perform O(log n) splits

Thus, an insertion in a (2,4) tree takes O(log n) time

Removal

We reduce deletion of an item to the case where the item is at the node with leaf children
Otherwise, we replace the item with its inorder successor (or, equivalently, with its inorder predecessor) and delete the latter item
Example: to delete key 24, we replace it with 27 (inorder successor)

Deleting an item from a node v may cause an underflow, where node v becomes a 1-node with one child and no keys
To handle an underflow at node v with parent u, we consider two cases:
Case 1: the adjacent siblings of v are 2-nodes

Fusion operation:

we merge v with an adjacent sibling w
we move an item from u to the merged node v'

After a fusion, the underflow may propagate to the parent u

Case 2: an adjacent sibling w of v is a 3-node or a 4-node

Transfer operation:

we move a child of w to v
we move an item from u to v
we move an item from w to u

After a transfer, no underflow occurs

Analysis of Removal in a (2,4) Tree

Let T be a (2,4) tree with n items

Tree T has O(log n) height

In a deletion operation

We visit O(log n) nodes to locate the node from which to delete the item
We handle an underflow with a series of O(log n) fusions, followed by at most one transfer
Each fusion and transfer takes O(1) time

Thus, deleting an item from a (2,4) tree takes O(log n) time

B-Trees

9.5 Red-Black Trees

A red-black tree is a representation of a (2,4) tree by means of a binary tree whose nodes are colored red or black
In comparison with its associated (2,4) tree, a red-black tree has:

same logarithmic time performance
simpler implementation with a single node type

A red-black tree can also be defined as a binary search tree that satisfies the following properties:

Root Property: the root is black
External Property: every leaf is black
Internal Property: the children of a red node are black
Depth Property: all the leaves have the same black depth

Theorem: A red-black tree storing n items has height O(log n)
Proof:

The height of a red-black tree is at most twice the height of its associated (2,4) tree, which is O(log n)

The search algorithm for a red-black tree is the same as that for a binary search tree
By the above theorem, searching in a red-black tree takes O(log n) time

9.5.1 Update Operations

Insertion

To perform operation insertItem(k, e), we execute the insertion algorithm for binary search trees and color red the newly inserted node z unless it is the root

We preserve the root, external, and depth properties
If the parent v of z is black, we also preserve the internal property and we are done
If the parent v of z is red, we have a double red (i.e., a violation of the internal property), which requires a reorganization of the tree

Example where the insertion of 4 causes a double red:

Consider a double red with child z and parent v, and let w be the sibling of v:

	Case 1: w is black The double red is an incorrect replacement of a 4-node Restructuring: we change the 4-node replacement
	Case 2: w is red The double red corresponds to an overflow Recoloring: we perform the equivalent of a split operation in (2,4)-tree

A restructuring remedies a child-parent double red when the parent red node has a black sibling
It is equivalent to restoring the correct replacement of a 4-node
The internal property is restored and the other properties are preserved

There are four restructuring configurations depending on whether the double red nodes are left or right children

A recoloring remedies a child-parent double red when the parent red node has a red sibling
The parent v and its sibling w become black and the grandparent u becomes red, unless it is the root
It is equivalent to performing a split operation on (2,4)-tree on a 5-node
The double red violation may propagate to the grandparent u

Analysis of Insertion in a Red-Black Tree

Recall that a red-black tree has O(log n) height
Step 1 takes O(log n) time because we visit O(log n) nodes
Step 2 takes O(1) time
Step 3 takes O(log n) time because we perform

O(log n) recolorings, each taking O(1) time, and
at most one restructuring taking O(1) time

Thus, an insertion in a red-black tree takes O(log n) time

Removal

To perform operation remove(k), we first execute the deletion algorithm for binary search trees
Let v be the internal node removed, w the external node removed, and r the sibling of w

If either v or r was red, we color r black and we are done
If v and r were both black, we color r double black, which is a violation of the depth property requiring a reorganization of the tree

Example where the deletion of 8 causes a double black:

The algorithm for remedying a double black node r with sibling y considers three cases
Case 1: y is black and has a red child

We perform a restructuring, equivalent to a transfer, and we are done

Case 2: y is black and its children are both black

We perform a recoloring, equivalent to a fusion operation on (2,4)-tree, which may propagate up the double black violation

Case 3: y is red

We perform an adjustment, equivalent to choosing a different representation of a 3-node, after which either Case 1 or Case 2 applies

Deletion in a red-black tree takes O(log n) time

Red-Black Tree Reorganization

Insertion (remedy double red)

Red-black tree action	(2,4) tree action	result
restructuring	change of 4-node representation	double red removed
recoloring	split	double red removed or propagated up

Deletion (remedy double black)

Red-black tree action	(2,4) tree action	result
restructuring	transfer	double black removed
recoloring	fusion	double black removed or propagated up
adjustment	change of 3-node representation	restructuring or recoloring follows

Red/Black Tree - visualization

9.5.2 C++ Implementation
html-9.11 (RBItem)
html-9.12 (RBTree1)
html-9.13 (RBTree2)
html-9.14 (RBTree3)