CSE 326: Data Structures Binary Search Trees

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1 nnouncements (1/3/0) S 36: ata Structures inary Search Trees Steve Seitz Winter 0 HW # due now HW #3 out today, due at beginning of class next riday. Project due next Wed. night. Read hapter 4 1 Ts Seen So ar The ictionary T Stack Push Pop Queue nqueue equeue Priority Queue Insert eletemin Then there is decreasekey ata: a set of (key, value) pairs Operations: Insert (key, value) ind (key) Remove (key) insert(seitz,.) find(ericm6) ericm6 ric, Mcambridge, seitz Steve Seitz S ericm6 ric Mcambridge S 18 soyoung Soyoung Shin S 00 3 The ictionary T is also called the Map T 4

2 Modest ew Uses Implementations Sets ictionaries Networks Operating systems ompilers : Router tables : Page tables : Symbol tables Unsorted Linked-list Unsorted array insert find delete Probably the most widely used T! Sorted array 6 inary Trees inary Tree: Representation inary tree is a root left subtree (maybe empty) right subtree (maybe empty) Representation: left pointer ata right pointer G H left right pointer pointer left right left right pointer pointer pointer pointer left right left right left right pointer pointer pointer pointer pointer pointer I J 8

3 Tree Traversals traversal is an order for visiting all the nodes of a tree Three types: Pre-order: Root, left subtree, right subtree In-order: Left subtree, root, right subtree Post-order: Left subtree, right subtree, root + * 4 (an expression tree) Inorder Traversal void traverse(node t){ if (t!= NULL) traverse (t.left); process t.element; traverse (t.right); inary Tree: Special ases inary Tree: Some Numbers Recall: height of a tree = longest path from root to leaf. or binary tree of height h: max # of leaves: G G max # of nodes: omplete Tree H I Perfect Tree List Tree min # of leaves: ull Tree min # of nodes: 11

4 inary Tree: Some Numbers inary Search Tree ata Structure Recall: depth of a node = path length from node to root. Structural property onsider the space of all possible binary trees of N nodes. Sum up the depths of every node in that forest and divide by the number of nodes. This is the average depth over all nodes over all binary trees of size N. How big is it? What would the average depth be for a well-balanced tree? each node has children result: storage is small operations are simple Order property all keys in left subtree smaller than root s key all keys in right subtree larger than root s key result: easy to find any given key xample and ounter-xample ind in ST, Recursive 8 1 Node ind(object key, Node root) { if (root == NULL) return NULL; INRY SRH TRS? Runtime: if (key < root.key) return ind(key, root.left); else if (key > root.key) return ind(key, root.right); else return root; 1 16

5 ind in ST, Iterative onus: indmin/indmax Node ind(object key, Node root) { while (root!= NULL && root.key!= key) { if (key < root.key) root = root.left; else root = root.right; return root; 1 1 Runtime: ind minimum ind maximum Insert in ST uildtree for ST 1 1 Runtime: Insert(13) Insert(8) Insert(31) Insertions happen only at the leaves easy! 1 Suppose keys 1,, 3, 4,, 6,, 8, are inserted into an initially empty ST. If inserted in given order, what is the tree? What big-o runtime for this kind of sorted input? If inserted in reverse order, what is the tree? What big-o runtime for this kind of sorted input?

6 uildtree for ST Suppose keys 1,, 3, 4,, 6,, 8, are inserted into an initially empty ST. eletion in ST If inserted median first, then left median, right median, etc., what is the tree? What is the big-o runtime for this kind of sorted input? 1 1 Why might deletion be harder than insertion? 1 eletion eletion The Leaf ase Removing an item disrupts the tree structure. asic idea: find the node that is to be removed. Then fix the tree so that it is still a binary search tree. Three cases: node has no children (leaf node) node has one child node has two children elete(1)

7 eletion The One hild ase eletion The Two hild ase elete(1) elete() 1 What can we replace with? 6 eletion The Two hild ase Idea: Replace the deleted node with a value guaranteed to be between the two child subtrees inally Options: replaces succ from right subtree: findmin(t.right) pred from left subtree : findmax(t.left) Now delete the original node containing succ or pred Leaf or one child case easy! Original node containing gets deleted 8

8 alanced ST Observations ST: the shallower the better! or a ST with n nodes verage depth (averaged over all possible insertion orderings) is O(log n) Worst case maximum depth is O(n) Simple cases such as insert(1,, 3,..., n) lead to the worst case scenario Solution: Require a alance ondition that 1. ensures depth is O(log n) strong enough!. is easy to maintain not too strong!