Binary Search Trees CMPSC 465 Related to CLRS Chapter 12 and Epp Section 10.6

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1 Binary Search Trees CMPSC 465 Related to CLRS Chapter 12 and Epp Section 10.6 I. Review & Extension Recall that we defined the term dictionary last chapter. Search trees fit the definition of dictionary; we re able to insert keys, delete them, and look up keys. Binary search trees are a subset of the set of search trees, where each node has at most 2 children. Also, going with the generalities we defined, binary search trees store dynamic sets. Warm-Up Problem and Referesher Discussion: a. Build a binary search tree from 65, 92, 84, 22, 34, 91, 98, 72. b. What s the worst case that could happen for looking up a value in a binary search tree? c. What s the best? How does it happen? d. Perform a preorder traversal on the tree from (a). e. Perform an inorder traversal on the tree from (a). We represent a binary search tree T via a linked data structure: points to the root of the tree. Each node contains attributes o key o satellite data (probably) o left, which points to the left child o right, which points to the right child o p, which points to the parent (and so T.root.p = ) Keys in binary search trees satisfy the binary search tree property: If y is in the left subtree of x, then If y is in the right subtree of x, then This treatment of the basics of binary search trees is brief since it s covered at length in 360. See notes-01-trees.pdf in the graphs and trees notes from 360 or Section 10.6 of Epp to review if you need to or you haven t worked with trees before. Also, Chapter 12 of CRLS expresses some of these ideas in formal pseudocode if you d like to see that. Page 1 of 6 Prepared by D. Hogan referencing CLRS - Introduction to Algorithms (3rd ed.) for PSU CMPSC 465

2 II. Minimum and Maximum of a Tree Consider the following tree: Question: Where is the smallest key located? Question: Where is the largest key located? Question: How does a search for a minimum go? Here is formal pseudocode for finding the extreme values of a (sub)tree rooted at x: TREE-MINIMUM(x) while x.left NIL x = x.left return x TREE-MAXIMUM(x) while x.right NIL x = x.right return x III. Successors and Predecessors When we have a situation where all keys in a BST are distinct, we define the successor of node x to be the node y s.t. y.key is the smallest possible choice of key s.t. key > x.key. When x has the largest key in the tree, x s successor is defined to be NIL. Example: Consider the tree to the right below and find the successors of various nodes listed, noting where it falls in the tree structure: Page 2 of 6 Prepared by D. Hogan referencing CLRS - Introduction to Algorithms (3rd ed.) for PSU CMPSC 465

3 There are two cases: If node x has a non-empty right subtree, then x s successor is If node x has an empty right subtree, then we need to move up. As we move to the left, we visit keys. So we visit the parents of nodes we see until we move right. We ll find that x s successor y is the node that x is the predecessor of. (This is because x is the maximum in y s left subtree.) Formally, here is the specification of the algorithm that finds the successor of node x: TREE-SUCCESSOR(x) if x.right NIL return TREE-MINIMUM(x.right) y = x.p while y NIL and x == y.right x = y y = y.p return y As you might expect, TREE-PREDECESSOR is symmetric to this algorithm. It comes up in the homework. Finally, we should consider the running time. For both predecessor and successor IV. Transplant There are times when we ll need to reorganize a tree by replacing a subtree with another subtree. Say we re replacing the subtree rooted at u with the subtree rooted at v. To do this, we must: Make u s parent become v s parent o unless u is the root. Then v becomes the root. Give u s parent v as a child. Whether v becomes a left or right child depends on what it was to u. We leave v.left and v.right alone, leaving that up to the routine that requires transplantation. Here s pseudocode for a routine that replaces the subtree rooted at u with the subtree rooted at v in tree T: TRANSPLANT(T, u, v) if u.p = = NIL T.root = v else if u = u.p.left u.p.left = v else u.p.right = v if v NIL v.p = u.p Example: Using u as the subtree rooted at 7 and v as the subtree rooted at 3, run TRANSPLANT(T, u, v). Page 3 of 6 Prepared by D. Hogan referencing CLRS - Introduction to Algorithms (3rd ed.) for PSU CMPSC 465

4 V. Deletion You might imagine that what we just did is a step toward deleting a node, but in our example, we lost not only the 3, but also its children. When deleting, we generally want to remove one node, but not its children too. We need to do more. We ll consider another copy of the same tree and look at some of the nicer cases first. Question: Which nodes in the tree to the right can we simply delete without doing anything else? Why? Question: Say we want to delete 7. What do we need to do? Question: How about 18? Harder Question: Working from the original tree, what happens if we try same strategy to delete 6? 15? look at candidates for replacing 6: if we bring 3 up, we get three children. if we bring 7 up, we re good, but lucky. look at candidates for replacing 15: if we bring 6 up, we have three children. if we bring 18 up, we get three children As it turns out, when the node we want to delete has two children, we ll need to preserve the binary search tree property by finding a node that fits in the hole we make. If we re deleting z, z s successor y will do the job. What do we know about y? must be in z s right subtree, must have no left children. draw picture. Our goal is to replace z by y. There are two cases: Case where y is z s right child: Case where y lies in z s right subtree but is not the root: Page 4 of 6 Prepared by D. Hogan referencing CLRS - Introduction to Algorithms (3rd ed.) for PSU CMPSC 465

5 To help summarize, the nice cases for deleting z from T are If z has no children, just remove z. If z has one child, make that child take z s position and drag the child s subtree along. Here s the pseudocode for deleting node z from BST T: TREE-DELETE(T, z) if z.left == NIL TRANSPLANT(T, z, z.right) else if z.right == NIL TRANSPLANT(T, z, z.left) else y = TREE-MINIMUM(z.right) if y.p z TRANSPLANT(T, y, y.right) y.right = z.right y.right.p = y // z has no left child // z has just a left child // z has two children // y is z s successor // y is in z s right subtree but is not the subtree s root TRANSPLANT(T, z, y) y.left = z.left y.left.p = y // replace z by y Example: Consider the tree given here and delete each node specified: I G K B Page 5 of 6 Prepared by D. Hogan referencing CLRS - Introduction to Algorithms (3rd ed.) for PSU CMPSC 465

6 VI. Where Next? Insertion and search (and deletion too) in a binary search tree run in O(h) time, where h is the height of the tree. The biggest performance issue is that if a binary search tree becomes unbalanced, h is not optimal. So, we ll look at other tree structures that solve the issue of balance. Homework: CRLS Exercises , , , ; Starting from the tree on the last page, delete C, J, B, L. Page 6 of 6 Prepared by D. Hogan referencing CLRS - Introduction to Algorithms (3rd ed.) for PSU CMPSC 465