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Dynamic Dictionaries
• Primary Operations:
get(key) => search
put(key, element) => insert
remove(key) => delete
• Additional operations:
ascend()
get(index)
remove(index)
Complexity Of Dictionary Operations
get(), put() and remove()
Data Structure Worst Case Expected
Hash Table O(n) O(1)
Binary Search
Tree
O(n) O(log n)
Balanced
Binary Search
Tree
O(log n) O(log n)
n is number of elements in dictionary
Complexity Of Other Operations
ascend(), get(index), remove(index)
Data Structure ascend get and
remove
Hash Table O(D + n log n) O(D + n log n)
Indexed BST O(n) O(n)
Indexed
Balanced BST
O(n) O(log n)
D is number of buckets
The Operation put()
Put a pair whose key is 35.
Remove From A Degree 1 Node
Remove from a degree 1 node. key = 40
Remove From A Degree 1 Node (contd.)
Remove from a degree 1 node. key = 15
Remove From A Degree 2 Node
Remove from a degree 2 node. key = 10
Remove From A Degree 2 Node
Replace with largest key in left subtree (or
smallest in right subtree).
Remove From A Degree 2 Node
Largest key must be in a leaf or degree 1 node.
Another Remove From A Degree 2 Node
Remove from a degree 2 node. key = 20
Remove From A Degree 2 Node
Replace with largest in left subtree.
Remove From A Degree 2 Node
Replace with largest in left subtree.
Yet Other Operations
• Priority Queue Motivated Operations:
find max and/or min
remove max and/or min
initialize
meld
Max And/Or Min
• Follow rightmost path to max element.
• Follow leftmost path to min element.
• Search and/or remove => O(h) time.
Initialize
• Sort Initialize search tree. n elements.
Output in inorder (O(n)).
• Initialize must takeisn’t possible to sort faster than O(n log n) time, because it O(n log n).
10 Meld
O(log n) Height Trees
• Full binary trees.
Exist only when n = 2k^ –1.
• Complete binary trees.
Exist for all n.
Cannot insert/delete in O(log n) time.
15 +^1
Balanced Search Trees
• Height balanced.
WAVL (Adelson-Velsky and Landis) trees
• Weight Balanced.
• Degree Balanced.
W2-3 trees
W2-3-4 trees
Wred-black trees
WB-trees
AVL Tree
• binary tree
• for every node x, define its balance factor
balance factor of x = height of left subtree of x
– height of right subtree of x
• balance factor of every node x is – 1, 0 , or 1
Balance Factors
This is an AVL tree.
Nh, h > 1
• Both L and R are AVL trees.
• The height of one is h-1.
• The height of the other is h-2.
• The subtree whose height is h-1 has Nh-1 nodes.
• The subtree whose height is h-2 has Nh-2 nodes.
• So, Nh = Nh-1 + Nh-2 + 1.
L R
Fibonacci Numbers
• F 0 = 0, F 1 = 1.
• Fi = Fi-1 + Fi-2 , i > 1.
• N 0 = 0, N 1 = 1.
• Nh = Nh-1 + Nh-2 + 1, i > 1.
• Nh = Fh+2 – 1.
• Fi ~ φi/sqrt(5).
• φ = (1 + sqrt(5))/2.
AVL Search Tree
