lecture 13

lecture 13

• 1.
  CS 332: AlgorithmsRed-Black Trees
• 2.
  Review: Binary SearchTrees Binary Search Trees (BSTs) are an important data structure for dynamic sets In addition to satellite data, eleements have: key : an identifying field inducing a total ordering left : pointer to a left child (may be NULL) right : pointer to a right child (may be NULL) p : pointer to a parent node (NULL for root)
• 3.
  Review: Binary SearchTrees BST property: key[left(x)]  key[x]  key[right(x)] Example: F B H K D A
• 4.
  Review: Inorder TreeWalk An inorder walk prints the set in sorted order: TreeWalk(x) TreeWalk(left[x]); print(x); TreeWalk(right[x]); Easy to show by induction on the BST property Preorder tree walk : print root, then left, then right Postorder tree walk : print left, then right, then root
• 5.
  Review: BST SearchTreeSearch(x, k) if (x = NULL or k = key[x]) return x; if (k < key[x]) return TreeSearch(left[x], k); else return TreeSearch(right[x], k);
• 6.
  Review: BST Search(Iterative) IterativeTreeSearch(x, k) while (x != NULL and k != key[x]) if (k < key[x]) x = left[x]; else x = right[x]; return x;
• 7.
  Review: BST InsertAdds an element x to the tree so that the binary search tree property continues to hold The basic algorithm Like the search procedure above Insert x in place of NULL Use a “trailing pointer” to keep track of where you came from (like inserting into singly linked list) Like search, takes time O( h ), h = tree height
• 8.
  Review: Sorting WithBSTs Basic algorithm: Insert elements of unsorted array from 1.. n Do an inorder tree walk to print in sorted order Running time: Best case:  ( n lg n ) (it’s a comparison sort) Worst case: O(n 2 ) Average case: O( n lg n ) (it’s a quicksort!)
• 9.
  Review: Sorting WithBSTs Average case analysis It’s a form of quicksort! for i=1 to n TreeInsert(A[i]); InorderTreeWalk(root); 3 1 8 2 6 7 5 5 7 1 2 8 6 7 5 2 6 7 5 3 1 8 2 6 5 7
• 10.
  Review: More BSTOperations Minimum: Find leftmost node in tree Successor: x has a right subtree: successor is minimum node in right subtree x has no right subtree: successor is first ancestor of x whose left child is also ancestor of x Intuition: As long as you move to the left up the tree, you’re visiting smaller nodes. Predecessor: similar to successor
• 11.
  Review: More BSTOperations Delete: x has no children: Remove x x has one child: Splice out x x has two children: Swap x with successor Perform case 1 or 2 to delete it F B H K D A C Example: delete K or H or B
• 12.
  Red-Black Trees Red-blacktrees : Binary search trees augmented with node color Operations designed to guarantee that the height h = O(lg n ) First: describe the properties of red-black trees Then: prove that these guarantee h = O(lg n ) Finally: describe operations on red-black trees
• 13.
  Red-Black Properties The red-black properties : 1. Every node is either red or black 2. Every leaf (NULL pointer) is black Note: this means every “real” node has 2 children 3. If a node is red, both children are black Note: can’t have 2 consecutive reds on a path 4. Every path from node to descendent leaf contains the same number of black nodes 5. The root is always black
• 14.
  Red-Black Trees Putexample on board and verify properties: 1. Every node is either red or black 2. Every leaf (NULL pointer) is black 3. If a node is red, both children are black 4. Every path from node to descendent leaf contains the same number of black nodes 5. The root is always black black-height : # black nodes on path to leaf Label example with h and bh values
• 15.
  Height of Red-BlackTrees What is the minimum black-height of a node with height h? A: a height- h node has black-height  h /2 Theorem: A red-black tree with n internal nodes has height h  2 lg( n + 1) How do you suppose we’ll prove this?
• 16.
  RB Trees: ProvingHeight Bound Prove: n -node RB tree has height h  2 lg( n +1) Claim: A subtree rooted at a node x contains at least 2 bh( x ) - 1 internal nodes Proof by induction on height h Base step: x has height 0 (i.e., NULL leaf node) What is bh(x)?
• 17.
  RB Trees: ProvingHeight Bound Prove: n -node RB tree has height h  2 lg( n +1) Claim: A subtree rooted at a node x contains at least 2 bh( x ) - 1 internal nodes Proof by induction on height h Base step: x has height 0 (i.e., NULL leaf node) What is bh(x)? A: 0 So…subtree contains 2 bh( x ) - 1 = 2 0 - 1 = 0 internal nodes (TRUE)
• 18.
  RB Trees: ProvingHeight Bound Inductive proof that subtree at node x contains at least 2 bh( x ) - 1 internal nodes Inductive step: x has positive height and 2 children Each child has black-height of bh( x ) or bh( x )-1 (Why?) The height of a child = (height of x ) - 1 So the subtrees rooted at each child contain at least 2 bh( x ) - 1 - 1 internal nodes Thus subtree at x contains (2 bh( x ) - 1 - 1) + (2 bh( x ) - 1 - 1) + 1 = 2•2 bh( x )-1 - 1 = 2 bh( x ) - 1 nodes
• 19.
  RB Trees: ProvingHeight Bound Thus at the root of the red-black tree: n  2 bh( root ) - 1 (Why?) n  2 h /2 - 1 (Why?) lg( n+1 )  h /2 (Why?) h  2 lg( n + 1) (Why?) Thus h = O(lg n )
• 20.
  RB Trees: Worst-CaseTime So we’ve proved that a red-black tree has O(lg n ) height Corollary: These operations take O(lg n ) time: Minimum(), Maximum() Successor(), Predecessor() Search() Insert() and Delete(): Will also take O(lg n ) time But will need special care since they modify tree