Binary Search Trees1 Chapter 3, Sections 1 and 2: Binary Search Trees AVL Trees   

Binary Search Trees2 Ordered Dictionaries Keys are assumed to come from a total order. As opposed to the Ch. 2 stuff where the keys were not necessarily totally ordered. New operations: closestKeyBefore(k) closestElemBefore(k) closestKeyAfter(k) closestElemAfter(k)

Binary Search Trees3 Our Theme… All of the data structures in Chapter 3 are useful for implementing ordered dictionaries. Typically, we are looking at variations on an ordered tree Want searching to be efficient Need to maintain the properties that make searching efficient, so inserting and removing elements can be complicated

Binary Search Trees4 First, a general strategy and a simple (though not really efficient) implementation….

Binary Search Trees5 Binary Search Binary search performs operation findElement(k) on a dictionary implemented by means of an array-based sequence, sorted by key similar to looking through a phone book for a name at each step, the number of candidate items is halved terminates after O(log n) steps Example: findElement(7) m l h m l h m l h l  m  h

Binary Search Trees6 Lookup Table A lookup 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: findElement takes O(log n) time, using binary search insertItem takes O(n) time since in the worst case we have to shift n items to make room for the new item removeElement take O(n) time since in the worst case we have to shift n 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)

Binary Search Trees7 Now for the Data Structures Pretty much all variations on the binary search tree theme

Binary Search Trees8 Binary Search Tree 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 Trees9 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 NO_SUCH_KEY Example: findElement(4) Algorithm findElement(k, v) if T.isExternal (v) return NO_SUCH_KEY if k  key(v) return findElement(k, T.leftChild(v)) else if k  key(v) return element(v) else { k  key(v) } return findElement(k, T.rightChild(v))   

Binary Search Trees10 Insertion (Nice case) 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 (i.e. we add two external children to w) Example: insert    w w

Binary Search Trees11 Insertion (Not so nice case) If k is already in the tree, and let let w be the node at which the key k is first encountered We call TreeSearch(k.leftChild(w)) and recursively apply algorithm to node returned. Example: insert  w w 2  

Binary Search Trees12 Deletion 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 v w  

Binary Search Trees13 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 v w z v 2 Next node in inorder traversal leftmost node in subtree rooted at right child

Binary Search Trees14 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 Trees15 AVL Trees v z

Binary Search Trees16 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 can differ by at most 1. An example of an AVL tree where the heights are shown next to the nodes:

Binary Search Trees17 Height of an AVL Tree Fact: The height of an AVL tree storing n keys is O(log n). Proof: First, we’re trying to bound the height of a tree with n keys. Let’s come at this the reverse way, and find the lower bound on the minimum number of internal nodes, n(h), that a tree with height h must have. Claim that n(h) grows exponentially. From this it will be easy to show that height of tree with n nodes is O(log n). So, formally, let n(h) be the minimum number of internal nodes of an AVL tree of height h. We easily see that n(1) = 1 and n(2) = n(1) n(2)

Binary Search Trees18 Height of an AVL Tree For n > 2, an AVL tree of height h with minimal number of nodes 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). This indicates that n(h) at least doubles every time we increase h by two, which intuitively says that n(h) grows at least exponentially. Let’s show this… By induction, n(h) > 2n(h-2), n(h) > 4n(h-4), n(h) > 8n(n-6), …, which implies that n(h) > 2 i n(h-2i) This holds for any positive integer i such that h-2i is positive. So, choose i so that we get a base case. That is, we want i such that either h-2i = 1 or h-2i = 2. I.e. i = (h/2)-1 or i=(h/2)-(1/2). 3 4 n(1) n(2)

Binary Search Trees19 Height of an AVL Tree Taking logarithms: h < 2log n(h) +2 Thus the height of an AVL tree is O(log n) 3 4 n(1) n(2)

Binary Search Trees20 Insertion in an AVL Tree Insertion is as in a binary search tree (may violate AVL property!) Always done by expanding an external node. Example: w b=x a=y c=z before insertion of 54after insertion Note that only nodes on path from w to root can now be unbalanced

Binary Search Trees21 Restoring Balance We use a “search and repair” strategy Let z be first unbalanced node encountered on path from w up toward root Let y be child of z with higher height (not y must be an ancestor of w), and x be child of y with higher height (in case of tie, go with child that is ancestor of w)

Binary Search Trees22 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 b=y a=z c=x T0T0 T1T1 T2T2 T3T3 b=y a=z c=x T0T0 T1T1 T2T2 T3T3 c=y b=x a=z T0T0 T1T1 T2T2 T3T3 b=x c=ya=z T0T0 T1T1 T2T2 T3T3 case 1: single rotation (a left rotation about a) case 2: double rotation (a right rotation about c, then a left rotation about a) (other two cases are symmetrical)

Binary Search Trees23 Insertion Example, continued T 0 T 1 T 2 T 3 x y z unbalanced......balanced T 1

Binary Search Trees24 Restructuring (as Single Rotations) Single Rotations:

Binary Search Trees25 Restructuring (as Double Rotations) double rotations:

Binary Search Trees26 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 32after deletion

Binary Search Trees27 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 w c=x b=y a=z

Binary Search Trees28 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)