1. Binary Search Trees < > = 6 2 9 1 4 8 Dictionaries
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Binary Search Trees
< 6 2
> 9 1 4 = 8

2. Ordered Dictionaries Keys are assumed to come from a total order.
New operations:
closestKeyBefore(k)
closestElemBefore(k)
closestKeyAfter(k)
closestElemAfter(k)

3. Binary Search (§3.1.1)
Binary search performs operation findElement(k) on a dictionary implemented by means of an array-based sequence, sorted by key
similar to the high-low game
at each step, the number of candidate items is halved
terminates after O(log n) steps
Example: findElement(7) 1 3 4 5 7 8 9 11 14 16 18 19 l m h 1 3 4 5 7 8 9 11 14 16 18 19 l m h 1 3 4 5 7 8 9 11 14 16 18 19 l m h 1 3 4 5 7 8 9 11 14 16 18 19
l=m =h

4. Lookup Table (§3.1.1)
A lookup table is a dictionary implemented by means of a sorted sequence
We store the items of the dictionary in an array-based sequence, sorted by key
We use an external comparator for the keys
Performance:
findElement takes O(log n) time, using binary search
insertItem takes O(n) time since in the worst case we have to shift n/2 items to make room for the new item
removeElement take O(n) time since in the worst case we have to shift n/2 items to compact the items after the removal
The lookup table is effective only for dictionaries of small size or for dictionaries on which searches are the most common operations, while insertions and removals are rarely performed (e.g., credit card authorizations)

5. Binary Search Tree (§3.1.2)
A binary search tree is a binary tree storing keys (or key-element pairs) at its internal nodes and satisfying the following property:
Let u, v, and w be three nodes such that u is in the left subtree of v and w is in the right subtree of v. We have key(u)  key(v)  key(w)
External nodes do not store items
An inorder traversal of a binary search trees visits the keys in increasing order 6 9 2 4 1 8

6. Dictionaries
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Search (§3.1.3)
To search for a key k, we trace a downward path starting at the root
The next node visited depends on the outcome of the comparison of k with the key of the current node
If we reach a leaf, the key is not found and we return NO_SUCH_KEY
Example: findElement(4)
Algorithm findElement(k, v)
if T.isExternal (v)
return NO_SUCH_KEY
if k < key(v)
return findElement(k, T.leftChild(v))
else if k = key(v)
return element(v)
else { k > key(v) }
return findElement(k, T.rightChild(v))
< 6 2
> 9 1 4 = 8

7. Insertion (§3.1.4)
< 6
To perform operation insertItem(k, o), we search for key k
Assume k is not already in the tree, and let w be the leaf reached by the search
We insert k at node w and expand w into an internal node
Example: insert 5 2 9
> 1 4 8
> w 6 2 9 1 4 8 w 5

8. Deletion (§3.1.5) 6
To perform operation removeElement(k), we search for key k
Assume key k is in the tree, and let let v be the node storing k
If node v has a leaf child w, we remove v and w from the tree with operation removeAboveExternal(w)
Example: remove 4
< 2 9
> v 1 4 8 w 5 6 2 9 1 5 8

9. Deletion (cont.) 1 v
We consider the case where the key k to be removed is stored at a node v whose children are both internal
we find the internal node w that follows v in an inorder traversal
we copy key(w) into node v
we remove node w and its left child z (which must be a leaf) by means of operation removeAboveExternal(z)
Example: remove 3 3 2 8 6 9 w 5 z 1 v 5 2 8 6 9

10. Performance (§3.1.6)
Consider a dictionary with n items implemented by means of a binary search tree of height h
the space used is O(n)
methods findElement , insertItem and removeElement take O(h) time
The height h is O(n) in the worst case and O(log n) in the best case

11. Dictionaries
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(2,4) Trees 9
10 14

12. Multi-Way Search Tree
A multi-way search tree is an ordered tree such that
Each internal node has at least two children and stores d -1 key-element items (ki, oi), where d is the number of children
For a node with children v1 v2 … vd storing keys k1 k2 … kd-1
keys in the subtree of v1 are less than k1
keys in the subtree of vi are between ki-1 and ki (i = 2, …, d - 1)
keys in the subtree of vd are greater than kd-1
The leaves store no items and serve as placeholders 15 30

13. Multi-Way Inorder Traversal
We can extend the notion of inorder traversal from binary trees to multi-way search trees
Namely, we visit item (ki, oi) of node v between the recursive traversals of the subtrees of v rooted at children vi and vi + 1
An inorder traversal of a multi-way search tree visits the keys in increasing order 8 12 15 2 4 6 10 14 18 30 1 3 5 7 9 11 13 16 19 15 17

14. Multi-Way Searching Similar to search in a binary search tree
A each internal node with children v1 v2 … vd and keys k1 k2 … kd-1
k = ki (i = 1, …, d - 1): the search terminates successfully
k < k1: we continue the search in child v1
ki-1 < k < ki (i = 2, …, d - 1): we continue the search in child vi
k > kd-1: we continue the search in child vd
Reaching an external node terminates the search unsuccessfully
Example: search for 30 15 30

15. (2,4) Tree
A (2,4) tree (also called 2-4 tree or tree) is a multi-way search with the following properties
Node-Size Property: every internal node has at most four children
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
2 8 12 18

16. Height of a (2,4) Tree
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  … + 2h-1 = 2h - 1
Thus, h  log (n + 1)
Searching in a (2,4) tree with n items takes O(log n) time
depth
items 1 1 2
h-1
2h-1 h

17. Insertion
We insert a new item (k, o) 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 v
2 8 12 18 v
2 8 12 18

18. Overflow and Split
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 u u v v'
v" 12 18 12 18
27 30 35 v1 v2 v3 v4 v5 v1 v2 v3 v4 v5

19. Analysis of Insertion Algorithm insertItem(k, o)
1. We search for key k to locate the insertion node v
2. We add the new item (k, o) at node v
3. while overflow(v)
if isRoot(v)
create a new empty root above v
v  split(v)
Let T be a (2,4) tree with n items
Tree T 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 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

20. Deletion
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)
2 8 12 18
2 8 12 18

21. Underflow and Fusion u u w v v'
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 and move an item from u to the merged node v'
After a fusion, the underflow may propagate to the parent u u u
9 14 9 w v v' 10
10 14

22. Underflow and Transfer
To handle an underflow at node v with parent u, we consider two cases
Case 2: an adjacent sibling w of v is a 3-node or a 4-node
Transfer operation:
1. we move a child of w to v
2. we move an item from u to v
3. we move an item from w to u
After a transfer, no underflow occurs u u
4 9
4 8 w v w v 2
6 8 2 6 9

23. Analysis of Deletion 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

24. Implementing a Dictionary
Comparison of efficient dictionary implementations
Search
Insert
Delete
Notes
Hash Table
1 expected
no ordered dictionary methods
simple to implement
Skip List
log n high prob.
randomized insertion
(2,4) Tree
log n worst-case
complex to implement

25. Dictionaries
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Red-Black Trees 6 v 3 8 z 4

26. Outline and Reading From (2,4) trees to red-black trees (§3.3.3)
Definition
Height
Insertion
restructuring
recoloring
Deletion
adjustment

27. From (2,4) to 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 4
3 5 4 5 3 6 OR 3 5 2 7

28. Red-Black Tree
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 9 4 15 2 6 12 21 7

29. Height of a Red-Black Tree
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 binary search 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

30. Insertion
To perform operation insertItem(k, o), 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
Else (v 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: 6 6 v v 3 8 3 8 z z 4

31. Remedying 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 4 4 w v w v 2 7 2 7 z z 6 6

32. Restructuring
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 z 4 6 w v v 2 7 4 7 z w 6 2

33. Restructuring (cont.)
There are four restructuring configurations depending on whether the double red nodes are left or right children 6 4 2 6 2 4 2 6 4 2 6 4 4 2 6

34. Recoloring
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 on a 5-node
The double red violation may propagate to the grandparent u 4 4 w v w v 2 7 2 7 z z 6 6
… 4 … 2
6 7

35. Analysis of Insertion Algorithm insertItem(k, o)
1. We search for key k to locate the insertion node z
2. We add the new item (k, o) at node z and color z red
3. while doubleRed(z)
if isBlack(sibling(parent(z)))
z  restructure(z)
return
else { sibling(parent(z) is red }
z  recolor(z)
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

36. Deletion
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
Else (v and r were both black) we color r double black, which is a violation of the internal property requiring a reorganization of the tree
Example where the deletion of 8 causes a double black: 6 6 v r 3 8 3 r w 4 4

37. Remedying a Double Black
The algorithm for remedying a double black node w 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, 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

38. 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