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Fall 2021 Fang Yu Software Security Lab. Data Structures Dept. Management Information Systems, National Chengchi University Lecture 11 Search Trees Binary Search Trees, AVL trees, and Splay Trees 3 Binary Search Trees ¡ A binary search tree is a binary tree storing keys (or key-value ¡ An inorder traversal of a binary entries) at its internal nodes and search trees visits the keys in an satisfying the following increasing order 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. 6 We have key(u) ≤ key(v) ≤ key(w) 2 9 ¡ External nodes do not store items 1 4 8 4 Search Algorithm TreeSearch(k, v) ¡ To search for a key k, we trace if T.isExternal (v) a downward path starting at the return v root if k < key(v) return TreeSearch(k, T.left(v)) ¡ The next node visited depends else if k key(v) on the comparison of k with the = return v key of the current node else { k > key(v) } ¡ If we reach a leaf, the key is return TreeSearch(k, T.right(v)) not found < 6 ¡ Example: get(4): 2 9 > ¡ Call TreeSearch(4,root) 1 4 = 8 ¡ The algorithms for floorEntry and ceilingEntry are similar 5 Insertion 6 ¡ To perform operation put(k, o), < we search for key k (using 2 9 > TreeSearch) 1 4 8 > ¡ Assume k is not already in the tree, and let w be the leaf w reached by the search 6 ¡ We insert k at node w and expand w into an internal node 2 9 ¡ Example: insert 5 1 4 8 w 5 6 Deletion 6 ¡ To perform operation remove(k), < we search for key k 2 9 > ¡ Assume key k is in the tree, and 1 4 v 8 let v be the node storing k w 5 ¡ 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 6 parent 2 9 ¡ Example: remove 4 1 5 8 7 Deletion (cont.) 1 ¡ We consider the case where the v 3 key k to be removed is stored at a node v whose children are both 2 8 internal 6 9 ¡ we find the internal node w that w follows v in an inorder traversal 5 ¡ we copy key(w) into node v z ¡ we remove node w and its left child z (which must be a leaf) by 1 v means of operation 5 removeExternal(z) 2 8 ¡ Example: remove 3 6 9 8 Performance ¡ Consider n ordered set items implemented by means of a binary search tree of height h ¡ the space used is O(n) ¡ methods get, put and remove take O(h) time ¡ The height h is O(n) in the worst case and O(log n) in the best case We want a balanced binary tree! 9 AVL Tree Definition ¡ AVL trees are balanced 4 ¡ An AVL Tree is a binary 44 search tree such that for 2 3 every internal node v of T, 17 78 the heights of the children of 1 2 1 v can differ by at most 1 32 50 88 1 1 48 62 An example of an AVL tree where the heights are shown next to the nodes: n(2) 3 10 4 n(1) 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) ¡ 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) AVL Trees 11 Insertion ¡ Insertion is as in a binary search tree ¡ Always done by expanding an external node. ¡ Example: 44 44 c=z 17 78 17 78 a=y 32 50 88 32 50 88 48 62 48 62 b=x 54 w before insertion after insertion After Insertion ¡ All nodes along the path increase their height by 1 ¡ It may violate the AVL property 44 5 44 4 17 78 4 17 78 3 32 50 3 88 32 50 2 88 48 62 2 48 62 1 54 1 Search and repair ¡ Let z be the first violation node from the bottom along the path ¡ Let y be z’child with the higher height (y is 2 greater than its sibling) ¡ Let x be y’s child with the higher height ¡ We rebalance z by calling trinode restructuring method 14 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 a=z a=z are symmetrical) case 2: double rotation (a right rotation about c, c=y then a left rotation about a) b=y T0 T0 c=x b=x T3 T1 b=y b=x T1 T2 T2 T3 a=z c=x a=z c=y case 1: single rotation (a left rotation about a) T0 T1 T2 T3 T0 T1 T2 T3 AVL Trees Restructuring 15 (as Single Rotations) ¡ Single Rotations: a = z single r otation b = y b = y a = z c = x c = x T0 T3 T1 T3 T0 T1 T2 T2 c = z single rotation b = y b = y a = x c = z a = x T3 T3 T0 T2 T2 T1 T0 T AVL Trees 1 Restructuring 16 (as Double Rotations) ¡ double rotations: a = z double rotation b = x c = y a = z c = y b = x T0 T2 T2 T3 T0 T1 T3 T1 c = z double rotation b = x a = y a = y c = z b = x T0 T2 T3 T2 T3 T1 T0 T1 AVL Trees 17 Insertion Example, continued 5 44 2 z 64 17 78 7 2 y 1 3 1 32 1 50 4 88 2 1 x 48 62 1 3 5 54 T3 unbalanced... T T0 2 4 T 1 44 4 2 3 x 17 62 6 2 y z 1 2 2 32 1 50 3 78 7 1 1 5 1 ...balanced 48 54 88 T 2 T0 T 1 T3 AVL Trees 18 Removal ¡ 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 unbalance. ¡ Example: 44 44 17 62 17 62 32 50 78 50 78 48 54 88 48 54 88 before deletion of 32 after deletion AVL Trees 19 Rebalancing after a 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 ¡ 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 AVL Trees 20 AVL Tree Performance ¡ a single restructure takes O(1) time ¡ using a linked-structure binary tree ¡ get takes O(log n) time ¡ height of tree is O(log n), no restructures needed ¡ put takes O(log n) time ¡ initial find is O(log n) ¡ Restructuring up the tree, maintaining heights is O(log n) ¡ remove takes O(log n) time ¡ initial find is O(log n) ¡ Restructuring up the tree, maintaining heights is O(log n) 21 Splay Tree ¡ a splay tree is a binary search tree where a node is splayed after it is accessed (for a search or update) ¡ deepest internal node accessed is splayed ¡ splay: move the node to the root ¡ splaying costs O(h), where h is height of the tree – which is still O(n) worst-case ¡ O(h) rotations, each of which is O(1) 22 Splay Tree ¡ which nodes are splayed after each operation? method splay node if key found, use that node get(k) if key not found, use parent of ending external node put(k,v) use the new node containing the entry inserted use the parent of the internal node that was actually removed from the tree (the parent of the node that the removed item was remove(k) swapped with) Searching in a Splay Tree: 23 Starts the Same as in a BST (20,Z) (10,A) (35,R) ¡ Search proceeds down the tree (14,J) to found item or an external (7,T) (21,O) (37,P) node. (1,Q) (8,N) (36,L) (40,X) ¡ Example: Search for the item with key 11. (1,C) (5,H) (7,P) (10,U) (2,R) (5,G) (5,I) (6,Y) Splay Trees 24 Example Searching in a BST, continued (20,Z) (10,A) (35,R) ¡ search for key 8, ends at an internal node.
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