Chapter 10 Search Trees Objectives Binary Search Trees
Chapter 10: Search Trees Objectives: • Binary Search Trees: Search, update, and implementation • AVL Trees: Properties and maintenance • 2 -4 Trees: Properties and maintenance • Red-black Trees: Properties and equivalence to 2 -4 Trees CSC 311: Data Structures 1
Ordered Dictionaries Keys are assumed to come from a total order. New operations: – first(): first entry in the dictionary ordering – last(): last entry in the dictionary ordering – successors(k): iterator of entries with keys greater than or equal to k; increasing order – predecessors(k): iterator of entries with keys less than or equal to k; decreasing order Search Trees CSC 311: Data Structures 2
Binary Search Binary search can perform operation find(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: find(7) 0 1 3 4 5 7 1 0 3 4 5 m l 0 9 11 14 16 18 m l 0 8 1 1 3 3 7 19 h 8 9 11 14 16 18 19 h 4 5 7 l m h 4 5 7 l=m =h Search Trees CSC 311: Data Structures 3
Search Tables A search 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: – – find takes O(log n) time, using binary search insert takes O(n) time since in the worst case we have to shift n/2 items to make room for the new item – remove 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) Search Trees CSC 311: Data Structures 4
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) An inorder traversal of a binary search trees visits the keys in increasing order 6 2 1 9 4 8 External nodes do not store items Search Trees CSC 311: Data Structures 5
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 nukk Example: find(4): Algorithm Tree. Search(k, v) if T. is. External (v) return v if k < key(v) return Tree. Search(k, T. left(v)) else if k = key(v) return v else { k > key(v) } return Tree. Search(k, T. right(v)) < – Call Tree. Search(4, root) 2 1 Search Trees CSC 311: Data Structures 6 9 > 4 = 8 6
Insertion To perform operation inser(k, o), we search for key k (using Tree. Search) 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 Search Trees 6 < 2 9 > 1 4 8 > w 6 2 1 CSC 311: Data Structures 9 4 8 5 w 7
Deletion 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 remove. External(w), which removes w and its parent Example: remove 4 Search Trees 6 < 2 9 > 4 v 1 w 8 5 6 2 1 CSC 311: Data Structures 9 5 8 8
Deletion (cont. ) 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 remove. External(z) 1 3 2 8 6 w 9 5 z 1 Example: remove 3 Search Trees v 5 v 2 8 6 CSC 311: Data Structures 9 9
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 Search Trees CSC 311: Data Structures 10
AVL Tree Definition 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 An example of an AVL tree where can differ by at the heights are shown next to the most 1. nodes. Search Trees CSC 311: Data Structures 11
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) > 2 n(h-2). So n(h) > 2 n(h-2), n(h) > 4 n(h-4), n(h) > 8 n(n-6), … (by induction), n(h) > 2 in(h-2 i) Solving the base case we get: n(h) > 2 h/2 -1 Taking logarithms: h < 2 log n(h) +2 Thus the height of an AVL tree is O(log n) Search Trees CSC 311: Data Structures 12
Insertion in an AVL Tree Insertion is as in a binary search tree Always done by expanding an external node. Example: 44 44 17 78 c=z a=y 32 50 48 88 62 32 50 48 w before insertion Search Trees CSC 311: Data Structures 88 62 b=x 54 after insertion 13
Trinode Restructuring let (a, b, c) be an inorder listing of x, y, z perform the rotations needed to make b the topmost node of the three (other two cases are symmetrical) a=z c=y b=y T 0 b=x c=x T 1 T 3 b=y T 2 T 1 T 3 case 1: single rotation (a left rotation about a) Search Trees case 2: double rotation (a right rotation about c, then a left rotation about a) a=z T 0 b=x T 2 c=x T 1 T 2 a=z T 3 CSC 311: Data Structures T 0 c=y T 1 T 2 T 3 14
Insertion Example, continued unbalanced. . . T 1 44 2 4 x 3 17 32 2 1 1 . . . balanced 48 62 y z 78 50 2 1 1 54 88 T 2 T 0 Search Trees CSC 311: Data Structures T 1 T 3 15
Restructuring: Single Rotations: Search Trees CSC 311: Data Structures 16
Restructuring: Double Rotations double rotations: Search Trees CSC 311: Data Structures 17
Removal in an AVL Tree Removal begins as in a binary search tree, which means the node removed will become an empty external node. Its parent, w, may cause an imbalance. Example: 44 44 17 62 32 50 48 17 78 54 50 88 before deletion of 32 Search Trees 62 CSC 311: Data Structures 48 78 54 88 after deletion 18
Rebalancing after Removal Let z be the first unbalanced node encountered while travelling up the tree from w. Also, let y be the child of z with the larger height, and let x be the child of y with the larger height. 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 a=z w 62 44 17 50 48 Search Trees c=x 78 54 44 b=y 62 17 50 48 88 CSC 311: Data Structures 78 88 54 19
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) – initial find is O(log n) – Restructuring up the tree, maintaining heights is O(log n) Search Trees CSC 311: Data Structures 20
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