Search Trees (Ch. 9) > = Binary Search Trees 1
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1 Search Trees (Ch. 9) < 6 > = Binary Search Trees 1
2 Ordered Dictionaries Keys are assumed to come from a total order. New operations: closestbefore(k) closestafter(k) Binary Search Trees
3 Binary Search Tree ( 9.1) A binary search tree is a binary tree storing keys (or key-element pairs) 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) External nodes do not store items An inorder traversal of a binary search trees visits the keys in increasing order Binary Search Trees 3
4 Search ( 9.1.1) 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 a null position Example: find(4) Algorithm find (k, v) if T.isExternal (v) return Position(null) if k < key(v) return find(k, T.leftChild(v)) else if k = key(v) return Position(v) else { k > key(v) } return find(k, T.rightChild(v)) < > = 9 Binary Search Trees 4
5 Insertion ( 9.1.) To perform operation insertitem(k, o), we search for key k Assume k is not already in the tree, and let 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 < > > w w 9 9 Binary Search Trees 5
6 Exercise: Binary Search Trees Insert into an initially empty binary search tree items with the following keys (in this order). Draw the resulting binary search tree 30, 40, 4, 58, 48, 6, 11, 13 Binary Search Trees 6
7 Deletion ( 9.1.) To perform operation removeelement(k), we search for key k Assume key k is in the tree, and let 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 removeaboveexternal(w) Example: remove w > < v Binary Search Trees 7
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 removeaboveexternal(z) Example: remove z 3 w 5 v 5 v Binary Search Trees 8
9 Exercise: Binary Search Trees Insert into an initially empty binary search tree items with the following keys (in this order). Draw the resulting binary search tree 30, 40, 4, 58, 48, 6, 11, 13 Now, remove the item with key 30. Draw the resulting tree Now remove the item with key 48. Draw the resulting tree. Binary Search Trees 9
10 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 findelement(), insertitem() and removeelement() take O(h) time The height h is O(n) in the worst case and O(log n) in the best case Binary Search Trees 10
11 AVL Trees v z Binary Search Trees 11
12 AVL Tree Definition ( 9.) 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 An example of an AVL tree where the heights are shown next to the nodes: Binary Search Trees 1
13 n() 3 Height of an AVL Tree 4 n(1) 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() = For n >, an AVL tree of height h contains the root node, one AVL subtree of height h-1 and another of height h-. That is, n(h) = 1 + n(h-1) + n(h-) Knowing n(h-1) > n(h-), we get n(h) > n(h-). So n(h) > n(h-), n(h) > 4n(h-4), n(h) > 8n(n-6), (by induction), n(h) > i n(h-i) Solving the base case we get: n(h) > h/-1 Taking logarithms: h < log n(h) + Thus the height of an AVL tree is O(log n) Binary Search Trees 13
14 Insertion in an AVL Tree Insertion is as in a binary search tree Always done by expanding an external node. Example: a=y c=z b=x w 54 before insertion after insertion Binary Search Trees 14
15 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 a=z b=y (other two cases are symmetrical) a=z c=y case : double rotation (a right rotation about c, then a left rotation about a) T 0 c=x T 0 b=x T 1 b=y T 3 b=x T T 3 a=z c=x T 1 T a=z c=y case 1: single rotation (a left rotation about a) T 0 T 1 T T 3 T 0 T 1 T T 3 Binary Search Trees 15
16 Insertion Example, continued 5 44 z y x unbalanced......balanced 54 5 T T 0 T T y x T z T 0 T 1 T 3 Binary Search Trees 16
17 Restructuring (as Single Rotations) Single Rotations: a = z b = y c = x single rotation a = z b = y c = x T 0 T 1 T T 3 T 0 T 1 T T 3 a = x b = y c = z single rotation a = x b = y c = z T 3 T 3 T 0 T T T 1 T 0 T 1 Binary Search Trees 17
18 Restructuring (as Double Rotations) double rotations: a = z b = x c = y double rotation a = z b = x c = y T 0 T T 1 T 3 T 0 T 1 T T 3 a = y b = x c = z double rotation a = y b = x c = z T 0 T 3 T T 3 T 1 T 0 T 1 T Binary Search Trees 18
19 Exercise: AVL Trees Insert into an initially empty AVL tree items with the following keys (in this order). Draw the resulting AVL tree 30, 40, 4, 58, 48, 6, 11, 13 Binary Search Trees 19
20 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: before deletion of 3 after deletion Binary Search Trees 0
21 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. 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 44 6 w 17 6 b=y c=x Binary Search Trees 1
22 Exercise: AVL Trees Insert into an initially empty AVL tree items with the following keys (in this order). Draw the resulting AVL tree 30, 40, 4, 58, 48, 6, 11, 13 Now, remove the item with key 48. Draw the resulting tree Now, remove the item with key 58. Draw the resulting tree Binary Search Trees
23 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) Binary Search Trees 3
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