Binary Search Trees Cormen cap 12 Edition 3

Binary Search Trees Cormen (cap 12, Edition 3) Estruturas de Dados e seus Algoritmos (Cap 4)

Dictionary Data Structures Goal: Design a data structure to store a small set of keys S={k 1, k 2, . . , kn} from a large universe U. – It shall efficiently support – – Query(x): determine whether a key x is in S or not – Insert(x): Add x to the set S if x is not there – Delete(x): Remove x from S if x is there Additional Goals – Low memory consumption – Efficient construction 2

Dictionary Data Structures Linked Lists ! Query(x): O(n) time ! Insert(x): Insert at the beginning of the list: O(1) time ! Delete(x): Find and then remove, O(n) time ! Construction time: O(n) time ! Space consumption: O(n) 3

Binary Search Trees n A Binary search tree(BST) T for a list K= (k 1 < ··· < kn) of n keys is a rooted tree that satisfies the following properties: – It has n internal nodes and n+1 leaves – Each internal node of T is associated with a distinct key – [Search Property] If node v is associated with key k i then: ! all nodes at the left subtree of v are associated with keys smaller than ki ! all nodes at the right subtree of v are associated with keys larger than ki 4

How does the BST works? Search Property: x y x

How does the BST works? Search Property: x y x x z

Binary Search Trees n Binary Search Trees k 2 k 1 k 4 k 3 k 5 7

Binary Search Trees n Keys are elements from a totally ordered set U – U can be the set of integers – U can be the set of students from a university 8

Binary Search Trees n Additional Properties – The key with minimum value is stored in the leftmost node of the tree – The key with maximum value is stored in the righttmost node of the tree 9

Binary Search Trees n Basic Operations – Query(x): Determine whether x belongs to T or not – Insert(x): if x is not in T, then insert x in T. – Delete(x): If x in T, then remove x from T 10

BST: Query(x) Algorithm Query(x) If x = leaf then Return “element was not found” End If If x = root then Return “element was found” Else if x < root then search in the left subtree else search in the right subtree End If

Binary Search Trees n Query(x) k 2 k 4 k 1 k 3 k 5 12

Inseting a new key Add a new element in the tree at the correct position in order to keep the search property. Algorithm Insert(x, T) If x = root. key then Return ‘x is already in T’ End If If root(T) is a leaf then Associate the leaf with x Return End If If x < root. key then Insert (x, left tree of T) else Insert (x, right tree of T) End If

Inseting a new key Example: Insert(50), Insert(20), Insert(39), Insert(8), Insert(79), Insert(26) 50 20 8 79 39 26

Removing a node in a BST SITUATIONS: Removing a leaf Removing an internal node with a unique child Removing an internal node with two children

Removing a Leaf 6 8 2 1 4 3

Removing a Leaf 6 8 2 1 4 3 6 8 2 1 4

Removing a node in a BST SITUATIONS: Removing a leaf Removing an internal node with a unique child Removing an internal node with two children

Removing an internal node with a unique child It is necessary to correct the pointer, “jumping” the node: the only grandchild becomes the right (left) son.

Removing an internal node with a unique child 6 8 2 1 4 3

Removing an internal node with a unique child 6 8 2 1 4 3 6 8 2 1 3

Removing a node in a BST SITUATIONS: Removing a leaf Removing an internal node with a unique child Removing an internal node with two children

Removing an internal node with two children • Find the element which preceeds the element to be removed considering the ordering (this corresponds to remove the rightmost element from the left subtree) Switch the information of the node to be removed with the node found

Removing an internal node with two children 6 8 2 1 4 3

Removing an internal node with two children 6 8 2 1 4 3 4 8 2 1 6 3

Removing an internal node with two children 4 8 2 1 6 3

Binary Search Trees: Operations Complexity n Basic Operations – Query(x): Determine whether x belongs to T or not ! Number of operations = O( height of T) – Insert(x): if x is not in T, then insert x in T. ! Number of operations = O( height of T) – Delete(x): If x in T, then remove x from T ! Number of operations = O( height of T) – Max(T) and Min(T) ! Number of operations = O( height of T) 34

Binary Search Trees: Operations Complexity n n Basic Operations – Query(x): Determine whether x belongs to T or not ! Number of operations = O( height of T) – Insert(x): if x is not in T, then insert x in T. ! Number of operations = O( height of T) – Delete(x): If x in T, then remove x from T ! Number of operations = O( height of T) – Max(T) and Min(T) ! Number of operations = O( height of T) Shallow trees are desirable 35

Binary Search Trees: Construction • Simple Approach: let k 1 , …, kn be the set of key not necessarily ordered. Proceed as follows: insert( k 1 ), insert( k 2 ) , . . . , insert( kn ) 36

Example: 50, 20, 39, 8, 79, 26, 58, 15, 88, 4, 85, 96, 71, 42, 53. 50 20 8 4 79 39 15 26 58 42 53 88 71 85 96

Binary Search Trees: Construction • Simple Approach: let k 1 , …, kn be the set of key not necessarily ordered. Proceed as follows: insert( k 1 ), insert( k 2 ) , . . . , !The insert( kn ) structure has height O(n) if the set of keys is ordered. !For a random permutation of the n first integers, its expected height is O(log n) (Cormen, 12. 3) 38

Binary Search Trees: Construction • Simple Approach: let k 1 , …, kn be the set of key not necessarily ordered. Proceed as follows: • Sort the keys BST(1: n) root(T) ‘median key’ left(root) <- BST(1, n/2) right(root) <- BST(n/2+1, n) End 39

Relation between #nodes and height of a binary tree At each level the number of nodes duplicates, such that for a binary tree with height h we have at most: 20+ 21 + 22 +. . . + 2 h-1 = 2 h – 1 nodes

Relation between #nodes and height of a binary tree At each level the number of nodes duplicates, such that for a binary tree with height h we have at most: 20+ 21 + 22 +. . . + 2 h-1 = 2 h – 1 nodes Or equivalently: The height of every binary search tree with n nodes is at least log n

The tree may become unbalanced Remove: node 8 6 8 2 1 4 3 6 2 1 4 3

The tree may become unbalanced Remove: node 8 Remove node 1 6 8 2 1 4 3 6 2 1 4 3

The tree may become unbalanced Remove: node 8 Remove node 1 6 8 2 1 4 3 6 6 2 2 1 4 4 3 3

The tree may become unbalanced The binary tree may become degenerate after operations of insertion and remotion: becoming a list, for example.

Balanced Trees Cormen (cap 13, Edition 3) Estruturas de Dados e seus Algoritmos (Cap 5)

AVL TREES (Adelson-Velskii and Landis 1962) BST trees that maintain a reasonable balanced tree all the time. Key idea: if insertion or deletion get the tree out of balance then fix it immediately All operations insert, delete, … can be done on an AVL tree with N nodes in O(log N) time (worst case)

AVL TREES AVL Tree Property: It is a BST in which the heights of the left and right subtrees of the root differ by at most 1 and in which the right and left subtrees are also AVL trees Height: length of the longest path from the root to a leaf.

AVL TREES: Example: 44 4 2 17 78 1 3 2 32 88 50 1 48 62 1 1 An example of an AVL tree where the heights are shown next to the nodes:

AVL TREES Other Examples:

Relation between #nodes and height of na AVL tree Let r be the root of an AVL tree of height h Let Nh denote the minimum number of nodes in an AVL tree of height h

Relation between #nodes and height of na AVL tree Let r be the root of an AVL tree of height h Let Nh denote the minimum number of nodes in an AVL tree of height h T r Te Td

Relation between #nodes and height of na AVL tree Let r be the root of an AVL tree of height h Let Nh denote the minimum number of nodes in an AVL tree of height h T r Te h-1 Td

Relation between #nodes and height of na AVL tree Let r be the root of an AVL tree of height h Let Nh denote the minimum number of nodes in an AVL tree of height h T r Te h-1 Td h-1 ou h-2

Relation between #nodes and height of na AVL tree Let r be the root of an AVL tree of height h Let Nh denote the minimum number of nodes in an AVL tree of height h T r Te h-1 Nh ≥ 1 + Nh-2 Td h-1 ou h-2 It grows faster than Fibonacci series Nh ≥ 1. 618 h-2

Relation between #nodes and height of na AVL tree Let r be the root of an AVL tree of height h Let Nh denote the minimum number of nodes in an AVL tree of height h T r Te h-1 Nh ≥ 1 + Nh-2 Td h-1 ou h-2 It grows faster than Fibonacci series Nh ≥ 1. 618 h-2 Height of AVL Tree <= 1. 44 log N (N is the number of nodes)

Height of AVL Tree Height of the tree is O(log. N) Where N is the number of elements contained in the tree n This implies that tree search operations Query(), Max(), Min() take O(log. N) time. n

Insertion in an AVL Tree Insertion is as in a binary search tree (always done by expanding an external node)

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 48 88 62

Insertion in an AVL Tree Insertion is as in a binary search tree (always done by expanding an external node) Example: Insert node 54 44 17 78 32 50 48 88 62

Insertion in an AVL Tree Insertion is as in a binary search tree (always done by expanding an external node) Example: Insert node 54 44 44 17 17 78 78 32 32 50 50 88 48 48 88 62 62 54

Insertion in an AVL Tree Insertion is as in a binary search tree (always done by expanding an external node) Example: Insert node 54 44 17 17 78 78 32 32 50 50 88 48 48 88 62 62 54

Insertion in an AVL Tree Insertion is as in a binary search tree (always done by expanding an external node) Example: Insert node 54 44 17 17 78 78 32 32 50 88 48 48 3 50 62 62 54 88

Insertion in an AVL Tree Insertion is as in a binary search tree (always done by expanding an external node) Example: Insert node 54 44 17 17 78 78 32 32 50 88 48 48 3 50 62 62 54 88 1

Insertion in an AVL Tree Insertion is as in a binary search tree (always done by expanding an external node) Example: Insert node 54 44 17 17 78 78 32 32 50 88 48 48 3 50 62 62 Unbalanced!! 54 88 1

How does the AVL tree work? After insertion and deletion we will examine the tree structure and see if any node violates the AVL tree property n If the AVL property is violated at node x, it means the heights of left(x) and right(x) differ by exactly 2 If it does violate the property we can modify the tree structure using “rotations” to restore the AVL tree property

Rotations Two types of rotations Single rotations – two nodes are “rotated” Double rotations – three nodes are “rotated” n n

Localizing the problem Two principles: • • Imbalance will only occur on the path from the inserted/deleted node to the root (only these nodes have had their subtrees altered - local problem) Rebalancing should occur at the deepest unbalanced node (local solution too)

Single Rotation (Right): Case I • Rotate x with left child y • x and y satisfy the AVL property after the rotation

Single Rotation (Left): Case II • Rotate x with right child y • x and y satisfy the AVL property after the rotation

Single Rotation - Example h h+1 Tree is an AVL tree by definition.

Single Rotation - Example h h+2 Node 02 added Tree violates the AVL definition! Perform rotation.

Single Rotation - Example x y h h+1 h B A Tree has this form. C

Example – After Rotation y x A B Tree has this form. C

Single Rotation Sometimes a single rotation fails to solve the problem k 1 k 2 k 1 Z h+2 h k 2 h X h+2 X Y l In such cases, we need to use a double-rotation Z Y

Double Rotations: Case IV

Double Rotations h h+1 Delete node 94 Tree is an AVL tree by definition.

Double Rotations h+2 h AVL tree is violated.

Double Rotations x y z C A B 1 Tree has this form. B 2

After Double Rotations z y A Tree has this form x B 1 B 2 C

Insertion We keep the height of each node x to check the AVL properrty Part 1. Perform normal BST insertion Part 2. Check AVL property and restore the property if necessary. To check whether the AVL property persists we only need to check the nodes in the path from the new leaf to the root of the BST because the balance of the other nodes are not affected – – Check if node x is balanced using the identity Height(x) = 1 + max { Height (left(x)), Height(right(x) } – We should update the heights of the visited nodes in this process

Insertion: Part 2 Detailed For each x in the path from the inserted leaf towards the root. If the heights of left(x) and right(x) height differ at most by 1 Do ‘nothing’ Else we know that one of the subtrees of x has height h and the other h+2 If the height of left(x) is h+2 then – If the height of left(x)) is h+1, we single rotate with left child (case 1) – Otherwise, the height of right(left(x)) is h+1 and we double rotate with left child (case 3) Otherwise, the height of right(x) is h+2 – If the height of right(x)) is h+1, then we rotate with right child (case 2) – Otherwise, the height of left(right(x)) is h+1 and we double rotate with right child (case 4) Break For

Insertion: Correctness • Let x be the deepest node that does not satisfy the AVL property. • Assume that case 2 occurs (the new element is inserted in tree C) • x and y satisfy the property after the rotation. • The ancestors of x satisfy the property because the height(x) before the insertion is h+2 and height(y) after the rotation is also h+2

Insertion: Correctness • Let x be the deepest node that does not satisfy the AVL property. • Assume that case 2 occurs (the new element is inserted in tree C) • The nodes in the path between the new element and y also satisfy the AVL property due to the assumption that x is the deepest node for which the AVL property does not hold • Nodes that are not in the path from the root to the new element are not affected

Insertion: Correctness • Let x be the deepest node that does not satisfy the AVL property. • Assume that case 4 occurs (the new element is inserted in tree B 1) • x, y and z satisfy the property after the rotation • The ancestors of x are balanced after the rotation because the height of x is h+2 before the insertion and the height of z is h+2 after the rotation.

Insertion: Correctness • Let x be the deepest node that does not satisfy the AVL property. • Assume that case 4 occurs (the new element is inserted in tree B 1) • The remaining nodes in the path between the new element and x also satisfy the property due to the assumption that x is the deepest node that does not satisfy the AVL property • The nodes that are not in the path between the new element and x are not affected.

Insertion: Complexity • • The time complexity to perform a rotation is O(1) since we just update a few pointers The time complexity to find a node that violates the AVL property depends on the height of the tree, which is log(N)

Deletion Perform normal BST deletion Perform verification similar to those employed for the insertion to restore the tree property

Summary AVL Trees Maintains a Balanced Tree Modifies the insertion and deletion routine Performs single or double rotations to restore structure Guarantees that the height of the tree is O(logn) The guarantee directly implies that functions find(), min(), and max() will be performed in O(logn) n n

Other Balanced trees • Red Black Trees (Cormen Cap 13, Jayme cap 6) • 2 -3 Trees (Hopcroft)

Dictionary Problem: non uniform access probabilities n We want to keep a data structure to support a sequence of INSERT, QUERY, DELETE operations – Some elements are accessed much more often than others non -uniform access probabilities 104

Dictionary Problem: non uniform access probabilities Consider the following AVL Tree 44 17 78 32 50 48 88 62

Dictionary Problem: non uniform access probabilities Consider the following AVL Tree 44 17 78 32 50 48 88 62 Suppose we want to search for the following sequence of elements: 48, 48, 62, 62, 48, 62.

Dictionary Problem: non uniform access probabilities Consider the following AVL Tree 44 17 In this case, is this a good structure? 78 32 50 48 88 62 Suppose we want to search for the following sequence of elements: 48, 48, 62, 62, 48, 62.

Dictionary Problem: non uniform access probabilities Consider the following AVL Tree 48 This structure is much better! 32 17 44 62 50 78 88 Suppose we want to search for the following sequence of elements: 48, 48, 62, 62, 48, 62.

Dictionary Problem: non uniform access probabilities Application: Building Inverted indexes n Given a text T, we want to design an inverted index S for T, that is, a structure that maintains for every word x of T, the list of positions where x occurs. T Positions ALO MEU AMIGO 123456789. . . 12. . . ALO AMIGO MEU …. ALO AMIGO MEU 30 34 40 1, 4, 30 12, 34 9, 40 109

Dictionary Problem: non uniform access probabilities Application: Building Inverted indexes n Given a text T, we want to design an inverted index S for T, that is, a structure that maintains for every word x of T, the list of positions where x occurs. T Positions ALO MEU AMIGO 123456789. . . 12. . . ALO AMIGO MEU n …. ALO AMIGO MEU 30 34 40 1, 4, 30 12, 34 9, 40 We do not know the list of words beforehand; some words may occur much more frequently than others 110

Dictionary Problem: non uniform access probabilities Static Case: distribution access probability is known beforehand • Lists • Optimal Binary Search Trees Dynamic Case: distribution access probability is not known beforehand • Self Adjusted Lists • Self Adjusted Binary Search Trees • Splay Trees 111

Dictionary Problem with non uniform access probabilities Problem n Given sequence K = k 1 < k 2 <··· < kn of n sorted keys, with a search probability pi for each key ki. – n n We assume that we always search an element that belongs to K. This assumption can be easily removed. Want to design a data structure with minimum expected search cost. Actual cost = # of items examined. – For key ki , number of elements accessed before finding k i 112

Optimal Binary Search Trees Cormen (cap 15. 5, Edition 3) Estruturas de Dados e seus Algoritmos (Cap 4)

Dictionary Problem: non uniform access probabilities Approach 1: Linked lists n n n Put the elements with highest probabilities of being accessed at the beginning of the list Keys (1, 2, 3, 4, 5); p=(0. 1, 0. 3, 0. 2, 0. 05, 0. 15) Best possible linked list ! 2 3 5 1 4 ! Expected cost of accessing an element = 1 x 0. 3 + 2 x 0. 2 + 3 x 0. 15 + 4 x 0. 1 + 5 x 0. 05 114

Dictionary Problem with non uniform access probabilities Approach 2: Binary Search Tree n n Given sequence K = k 1 < k 2 <··· < kn of n sorted keys, with a search probability pi for each key ki. Want to build a binary search tree (BST) with minimum expected search cost. Actual cost = # of items examined. For key ki , cost = depth. T (ki) + 1, where depth. T(ki) = depth of ki in BST T. (root is at depth 0) 115

Expected Search Cost Sum of probabilities is 1. Identity (1) 116

Example Consider 5 keys with these search probabilities: p 1 = 0. 25, p 2 = 0. 2, p 3 = 0. 05, p 4 = 0. 2, p 5 = 0. 3. k 2 k 1 i 1 2 3 4 5 k 4 k 3 depth. T ( k i) 1 0 2 1 2 depth. T ( k i) · pi 0. 25 0 0. 1 0. 2 0. 6 1. 15 k 5 Therefore, E[search cost] = 2. 15. 117

Example p 1 = 0. 25, p 2 = 0. 2, p 3 = 0. 05, p 4 = 0. 2, p 5 = 0. 3. k 2 k 1 k 5 i depth. T(ki) 1 1 2 0 3 3 4 2 5 1 depth. T(ki)·pi 0. 25 0 0. 15 0. 4 0. 3 1. 10 k 4 Therefore, E[search cost] = 2. 10. k 3 This tree turns out to be optimal for this set of keys. 118

Example Observations: n n Optimal BST may not have smallest height. Optimal BST may not have highest-probability key at root. Build by exhaustive checking? n n n Construct each n-node BST. For each, assign keys and compute expected search cost. But there are (4 n/n 3/2) different BSTs with n nodes. 119

Optimal Substructure Any subtree of a BST contains keys in a contiguous range ki, . . . , kj for some 1 ≤ i ≤ j ≤ n. T T If T is an optimal BST and T contains subtree T ’ with keys ki, . . . , kj , then T must be an optimal BST for keys ki, . . . , kj. Proof: Otherwise, we can obtain a tree better T by replacing T’ with an optimal BST for keys ki, . . . , kj. 120

Optimal Substructure One of the keys in ki, …, kj, say kr, where i ≤ r ≤ j, must be the root of an optimal subtree for these keys. Left subtree of kr contains ki, . . . , kr 1. kr Right subtree of kr contains kr+1, . . . , kj. To find an optimal BST: n n ki kr-1 kr+1 kj Examine all candidate roots kr , for i ≤ r ≤ j Determine all optimal BSTs containing ki, . . . , kr 1 and containing kr+1, . . . , kj 121

Recursive Solution When the OPT subtree becomes a subtree of a node: Depth of every node in OPT subtree goes up by 1. Expected search cost increases by n n from Identity (1)

Recursive Solution When the OPT subtree becomes a subtree of a node: Depth of every node in OPT subtree goes up by 1. Expected search cost increases by n n from Identity (1) k 2 k 1 k 4 k 3 k 1 k 5 k 0 k 4 k 3 k 5 123

Recursive Solution e[i, j]: cost of the optimal BST for ki, . . , kj : If kr is the root of an optimal BST for ki, . . , kj : n e[i, j ] = pr + ( e[i, r 1] + w(i, r 1) ) + ( e[r+1, j] + w(r+1, j) )= e[i, r 1] + e[r+1, j] + w(i, j). But, we don’t know kr. Hence, 124

Computing an Optimal Solution For each subproblem (i, j), store: expected search cost in a table e [1. . n+1 , 0. . n] n Will use only entries e[i, j ], where j ≥ i 1. root[i, j ] = root of subtree with keys ki, . . , kj, for 1 ≤ i ≤ j ≤ n. w[1. . n+1, 0. . n] = sum of probabilities n n w[i, i 1] = 0 for 1 ≤ i ≤ n. w[i, j ] = w[i, j-1] + pj for 1 ≤ i ≤ j ≤ n. 125

Pseudo-code 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. OPTIMAL-BST(p, q, n) for i ← 1 to n + 1 do e[i, i 1] ← 0 Consider all trees with l keys. for len ← 1 to n Fix the first key. do for i ← 1 to n len + 1 Fix the last key do j ←i + len 1 e[i, j ]←∞ for r ←i to j do t ← e[i, r 1] + e[r + 1, j ] + w[i, j ] Determine the root if t < e[i, j ] of the optimal then e[i, j ] ← t (sub)tree root[i, j ] ←r return e and root Time: O(n 3) Space: O(n 2) 126

Speeding up the Algorithm Knuth principle: Let kr be the root of an optimal BST for the set of keys ki <. . < kj. Furthermore, let kj+1 > kj and ki-1<ki. Then, (i) there is an optimal BST for the set of keys ki-1 , ki, . . . , kj with root smaller than or equal to kr (ii) there is an optimal BST for the set of keys ki , ki+1, . . . , kj+1 with root larger than or equal to kr 127

Knuth principle: Example p 1 = 0. 25, p 2 = 0. 2, p 3 = 0. 05, p 4 = 0. 2, p 5 = 0. 3. k 2 k 1 k 5 k 4 k 3 • Let k 0 be a key with probability p 0 then there is an optimal BST for the set (k 0, …, k 5) with root smaller than or equal to k 2. 128

Knuth principle: Example p 1 = 0. 25, p 2 = 0. 2, p 3 = 0. 05, p 4 = 0. 2, p 5 = 0. 3. k 2 k 1 k 5 k 4 k 3 • Let k 6 be a key with probability p 6 then there is an optimal BST for the set (k 1, …, k 6) with root larger than or equal to k 2 129

Speeding up the Algorithm 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. OPTIMAL-BST-Revised(p, q, n) for i ← 1 to n + 1 do e[i, i 1] ← 0 Consider all trees with l keys. for len ← 1 to n O(n l ) do for i ← 1 to n len + 1 do j ←i + len 1 Optimization. e[i, j ]←∞ for r ←root[i, j-1] to root[i+1, j] do t ← e[i, r 1] + e[r + 1, j ] + w[i, j ] Determine the root if t < e[i, j ] of the optimal then e[i, j ] ← t (sub)tree root[i, j ] ←r return e and root Time: O(n 2) Space: O(n 2) 130

Speeding up the Algorithm 131

Lower Bound on the expected search cost 132

Lower Bound on the expected search cost 133

Lower Bound on the expected search cost 134