1. (2,4) Trees 9 2 5 7 10 14 5/6/2019 10:26 AM (2,4) Trees (2,4) Trees
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(2,4) Trees

2. Outline and Reading Multi-way search tree (§3.3.1) (2,4) tree (§3.3.2)
Definition
Search
(2,4) tree (§3.3.2)
Insertion
Deletion
Comparison of dictionary implementations
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(2,4) Trees

3. 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
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(2,4) Trees

4. 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
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(2,4) Trees

5. 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
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(2,4) Trees

6. (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
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(2,4) Trees

7. Height of a (2,4) Tree
Theorem: A (2,4) tree storing n items has height Θ(log n)
Proof:
Let h be the height of a (2,4) tree with n items
By depth property, we can have at most 4 nodes at depth 1, at most 42 at depth 2, etc. Thus number of external nodes is at most 4h
Also by depth property, have at least 2 nodes at depth 1, 22 nodes at depth 2, and in general at least 2h external nodes.
Fact: a multiway search tree storing n items has n+1 external nodes (homework: C-3.16)
Thus, 2h≤n+1 and n+1≤4h
Taking logs gives h≤log(n+1) and log(n+1) ≤2h which proves the theorem.
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(2,4) Trees

8. Height of a (2,4) Tree
Searching in a (2,4) tree with n items takes O(log n) time
A direct consequence of our height theorem. 1 2
2h-1
items
h-1 h
depth
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(2,4) Trees

9. 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
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(2,4) Trees

10. 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
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(2,4) Trees

11. 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 may perform O(log n) splits
Thus, an insertion in a (2,4) tree takes O(log n) time
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(2,4) Trees

12. 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)
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2 8 12 18
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(2,4) Trees

13. 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
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(2,4) Trees

14. Underflow and Transfer
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
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(2,4) Trees

15. 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
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(2,4) Trees

16. 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
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(2,4) Trees

17. Red-Black Trees v z 6 3 8 4 5/6/2019 10:26 AM (2,4) Trees (2,4) Trees

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

19. 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
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(2,4) Trees

20. 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
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(2,4) Trees

21. 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
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(2,4) Trees

22. 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
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(2,4) Trees

23. 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
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(2,4) Trees

24. 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
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(2,4) Trees

25. 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
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(2,4) Trees

26. 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
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27. 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
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(2,4) Trees

28. 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 of 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
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(2,4) Trees

29. 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
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(2,4) Trees

30. 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
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(2,4) Trees