1. Dictionaries
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ADT for Map:
Map stores elements (entries) so that they can be located quickly using keys.
Each element (entry) is a key-value pair (k, v), where k is the key and v can be any object to store additional information.
Each key is unique. (different entries have different keys.)
Map support the following methods:
Size(): Return the number of entries in M
isEmpty(): Test whether M is empty
get(k); If M contains an entry e with key=k, then return e else return null.
put(k, v): If M does not contain an entry with key=k then add (k, v) to the map and return null; else replace the entry with (k, v) and return the old value.
Binary Search Trees

2. Methods of Map (continued)
Dictionaries
二○一九年九月二十四日
Methods of Map (continued)
remove(k): remove from M the netry with key=k and return its value; if M has no such entry with key=k then return null.
keys(); Return an iterable collection containing all keys stored in M
values(): Return an iterable collection containing all values in M
entries(): return an iterable collection containing all key-value entries in M.
Remakrs: hash table is an implementation of Map.
Binary Search Trees

3. ADT for Dictionary: A Dictionary stores elements (entries).
Dictionaries
二○一九年九月二十四日
ADT for Dictionary:
A Dictionary stores elements (entries).
Each element (entry) is a key-value pair (k, v), where k is the key and v can be any object to store additional information.
The key is NOT unique.
Dictionary support the following methods:
size(): Return the number of entries in D
isEmpty(): Test whether D is empty
find(k): If D contains an entry e with key=k, then return e else return null.
findAll(k): Return an iterable collection containing all entries with key=k.
insert(k, v): Insert an entry into D, returning the entry created.
remove(e): remove from D an enty e, returing the removed entry or null if e was not in D.
entries(): return an iterable collection of the key-value entries in D.
Binary Search Trees

4. Part-F1 Binary Search Trees
Dictionaries
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Part-F1 Binary Search Trees
< 6 2
> 9 1 4 = 8
Binary Search Trees

5. Search Trees
Tree data structure that can be used to implement a dictionary.
find(k): If D contains an entry e with key=k, then return e else return null.
findAll(k): Return an iterable collection containing all entries with key=k.
insert(k, v): Insert an entry into D, returning the entry created.
remove(e): remove from D an enty e, returing the removed entry or null if e was not in D.
Binary Search Trees

6. Binary Search Trees
A binary search tree is a binary tree storing keys (or key-value entries) 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)
Different nodes can have the same key.
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
Binary Search Trees

7. Dictionaries
二○一九年九月二十四日
Search
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 null
Example: find(4):
Call TreeSearch(4,root)
Algorithm TreeSearch(k, v)
if T.isExternal (v)
return v
if k < key(v)
return TreeSearch(k, T.left(v))
else if k = key(v)
else { k > key(v) }
return TreeSearch(k, T.right(v))
< 6 2
> 9 1 4 = 8
Binary Search Trees

8. Insertion < > > w w 6 2 9 1 4 8 6 2 9 1 4 8 5
To perform operation insert(k, o), we search for key k (using TreeSearch)
Algorithm TreeINsert(k, x, v):
Input: A search key, an associate value x and a node v of T to start with
Output: a new node w in the subtree T(v) that stores the entry (k, x)
W TreeSearch(k,v)
If k=key(w) then
return TreeInsert(k, x, T.left(w))
T.insertAtExternal(w, (k, x))
Return
Example: insert 5
Example: insert another 5? 2 9
> 1 4 8
> w 6 2 9 1 4 8 w 5
Binary Search Trees

9. Deletion To perform operation remove(k), we search for key k <6
To perform operation remove(k), we search for key k
Assume key k is in the tree, and 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 removeExternal(w), which removes w and its parent and replace v with the remaining child.
Example: remove 4
< 2 9
> v 1 4 8 w 5 6 2 9 1 5 8
Binary Search Trees

10. 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 removeExternal(z)
Example: remove 3 3 2 8 6 9 w 5 z 1 v 5 2 8 6 9
Binary Search Trees

11. Deletion (Another Example)1 v 3 2 8 6 9 w 4 z 5 1 v 4 2 8 6 9 5
Binary Search Trees

12. Later, we will try to keep h =O(log n).
Performance
Consider a dictionary with n items implemented by means of a binary search tree of height h
the space used is O(n)
methods find, insert and remove take O(h) time
The height h is O(n) in the worst case and O(log n) in the best case
Later, we will try to keep h =O(log n).
Review the past
Binary Search Trees

13. Part-F2 AVL Trees v z 6 3 8 4 Binary Search Trees Dictionaries
二○一九年九月二十四日
Part-F2 AVL Trees 6 3 8 4 v z
Binary Search Trees

14. AVL Tree Definition (§ 9.2)
AVL trees are balanced.
An AVL Tree is a binary search tree such that for every internal node v of T, the heights of the children of v can differ by at most 1.
An example of an AVL tree where the heights are shown next to the nodes:
Binary Search Trees

15. Balanced nodes
A internal node is balanced if the heights of its two children differ by at most 1.
Otherwise, such an internal node is unbalanced.
Binary Search Trees

16. 3 4
n(1)
n(2)
Height of an AVL Tree
Fact: The height of an AVL tree storing n keys is O(log n).
Proof: Let us bound n(h): the minimum number of internal nodes of an AVL tree of height h.
We easily see that n(1) = 1 and n(2) = 2
For n > 2, an AVL tree of height h contains the root node, one AVL subtree of height n-1 and another of height n-2.
That is, n(h) = 1 + n(h-1) + n(h-2)
Knowing n(h-1) > n(h-2), we get n(h) > 2n(h-2). So
n(h) > 2n(h-2), n(h) > 4n(h-4), n(h) > 8n(n-6), … (by induction),
n(h) > 2in(h-2i)>2 {h/2 -1} (1) = 2 {h/2 -1}
Solving the base case we get: n(h) > 2 h/2-1
Taking logarithms: h < 2log n(h) +2
Thus the height of an AVL tree is O(log n)
h-1
Binary Search Trees
h-2

17. Insertion in an AVL Tree
Insertion is as in a binary search tree
Always done by expanding an external node.
Example: 44 17 78 32 50 88 48 62 44 17 78 32 50 88 48 62 54
c=z
a=y
b=x w
before insertion
after insertion
It is no longer balanced
Binary Search Trees

18. Names of important nodes
w: the newly inserted node. (insertion process follow the binary search tree method)
The heights of some nodes in T might be increased after inserting a node.
Those nodes must be on the path from w to the root.
Other nodes are not effected.
z: the first node we encounter in going up from w toward the root such that z is unbalanced.
y: the child of z with higher height.
y must be an ancestor of w. (why? Because z in unbalanced after inserting w)
x: the child of y with higher height.
x must be an ancestor of w.
The height of the sibling of x is smaller than that of x. (Otherwise, the height of y cannot be increased.)
See the figure in the last slide.
Binary Search Trees

19. Algorithm restructure(x):
Input: A node x of a binary search tree T that has both parent y and grand-parent z.
Output: Tree T after a trinode restructuring.
Let (a, b, c) be the list (increasing order) of nodes x, y, and z. Let T0, T1, T2 T3 be a left-to-right (inorder) listing of the four subtrees of x, y, and z not rooted at x, y, or z.
Replace the subtree rooted at z with a new subtree rooted at b..
Let a be the left child of b and let T0 and T1 be the left and right subtrees of a, respectively.
Let c be the right child of b and let T2 and T3 be the left and right subtrees of c, respectively.
Binary Search Trees

20. Restructuring (as Single Rotations)3 2 1
a = x
b = y
c = z
single rotation
Binary Search Trees

21. Restructuring (as Double Rotations)
c = z
b = x
a = y T 3 1 2
Binary Search Trees

22. Insertion Example, continued
unbalanced... 4 T 1 44 x 2 3 17 62 y z 1 2 2 32 50 78 1 1 1
...balanced 48 54 88 T 2 T T 1 T
Binary Search Trees 3

23. Theorem:
One restructure operation is enough to ensure that the whole tree is balanced.
Proof: Look at the four cases on slides 20 and 21.
Binary Search Trees

24. Removal in an AVL Tree
Removal begins as in a binary search tree by calling removal(k) for binary tree.
may cause an imbalance.
Example: 44 17 78 32 50 88 48 62 54 44 w 17 62 50 78 48 54 88
before deletion of 32
after deletion
Binary Search Trees

25. Rebalancing after a Removal
Let z be the first unbalanced node encountered while travelling up the tree from w. w-parent of the removed node (in terms of structure, not the name)
let y be the child of z with the larger height,
let x be the child of y defined as follows;
If one of the children of y is taller than the other, choose x as the taller child of y.
If both children of y have the same height, select x be the child of y on the same side as y (i.e., if y is the left child of z, then x is the left child of y; and if y is the right child of z then x is the right child of y.)
The way to obtain x, y and z are different from insertion.
Binary Search Trees

26. Rebalancing after a Removal
We perform restructure(x) to restore balance at z.
As this restructuring may upset the balance of another node higher in the tree, we must continue checking for balance until the root of T is reached 62
a=z 44 44 78 w 17 62
b=y 17 50 88 50 78
c=x 48 54 48 54 88
Binary Search Trees

27. Unbalanced after restructuring1 1 62
h=3
a=z 44
h=4
h=5
h=5 44 78 w 17 62
b=y 17 50 88 32 50 78
c=x 88
Binary Search Trees

28. Rebalancing after a Removal
We perform restructure(x) to restore balance at z.
As this restructuring may upset the balance of another node higher in the tree, we must continue checking for balance until the root of T is reached 62
a=z 44 44 78 w 17 62
b=y 17 50 88 50 78
c=x 48 54 48 54 88
Binary Search Trees

29. Example a: Which node is w? Let us remove node 17. w44 17 78 32 50 88 48 62 54 44 w 32 62 50 78 48 54 88
before deletion of 32
after deletion
Binary Search Trees

30. Rebalancing: We perform restructure(x) to restore balance at z.
As this restructuring may upset the balance of another node higher in the tree, we must continue checking for balance until the root of T is reached 62
a=z 44 44 78 w 32 62
b=y 32 50 88 50 78
c=x 48 54 48 54 88
Binary Search Trees

31. Running Times for AVL Trees
a single restructure is O(1)
using a linked-structure binary tree
find is O(log n)
height of tree is O(log n), no restructures needed
insert is O(log n)
initial find is O(log n)
Restructuring up the tree, maintaining heights is O(log n)
remove is O(log n)
Binary Search Trees