CMSC 441: Algorithms. Hillol Kargupta, Professor.

Transcription

1 CMSC 441: Algorithms Hillol Kargupta, Professor

2 Today s Topics Binary Search Trees Red-Black Trees

3 Binary Search Tree Binary tree Satisfies binary-search-tree property: Let x be a node. If y is a node in the left subtree of x, then key[y]<=key[x]. If y is a node in the right subtree of x, then key[x]<=key[y]. 3

4 Traversing a Binary Search Tree Inorder left[x], x, right[x] Preorder x, left[x], right[x] Postorder Left[x], right[x], x 4

5 Tree Search Tree-Search(x, k) If x = IL or k=key[x] Then return x If k < key[x] Then return Tree-Search(left[x], k) Else return Tree-Search(right[x], k) 5

6 Min, Max, and Successor Left most node Right most node Successor of a node x is the smallest key greater than key[x]. 6

7 Successor Tree-Successor(x) If right[x]!= IL Then return Tree-Min(right[x]) Y=p[x] While y!= IL and x = right[y] Do x=y Y = p[y] Return y 7

8 Observation If a node in a binary search tree has two children then its successor has no left child. 8

9 Predecessor in BST Predecessor of a node v in a BST is the node that holds the data value that immediately precedes the data at v in order. Finding predecessor v has a left subtree then predecessor must be the largest value in the left subtree (the rightmost node in the left subtree) v does not have a left subtree predecessor is the first node on path back to root that does not have v in its left subtree 8/3/007

10 Successor in BST Successor of a node v in a BST is the node that holds the data value that immediately follows the data at v in order. Finding Successor v has right subtree successor is smallest value in right subtree (the leftmost node in the right subtree) v does not have right subtree successor is first node on path back to root that does not have v in its right subtree

11 Average Case Analysis Internal Path Length, D() =sum of the depths of all nodes in a tree with nodes Consider a tree with the left subtree with i nodes and right subtree with -i-1 nodes D()=D(i)+D(-i-1)+-1 Replacing D(i) and D( - i -1) by the average internal D( ) D(1) 0 1 j0 D( j) 1 path length

12 ( Proof Sketch Replacing ( -1) by c where c is some constant 1 D( ) D( j) c D( ) j0 1 j0 D( j) c Substituting -1for in Equation 1, 1) D( 1) j0 D( j) c( Equation 1. 1) Equation. Subtracting Equation from Equation 1, D( ) ( 1) D( 1) c Equation 3

13 Continued 3 (1) 3 () ) ( 1) ( 1 1) ( 1 ) ( 1) Equation 3 by ( Dividing both sides of c D D c D D c D D

14 Continued ) (lg O( lg ) Average height ) lg ( ) ( ) (lg 1) ( ) ( 1 1) ( ) ( 1 (1) 1 ) ( the equations in the previous slide, Adding all O O D O c D i c D i c D D i i

15 Insertion Must maintain binary search tree property T(n) = O(h) where h is the height of the BST 15

16 Deletion Tree-Delete must maintain binary search tree property Considers three cases: z has no children Modify parent s child pointer z has one child Modify child s parent pointer Modify parent s child pointer z has two children Replace z with it s successor Modify pointers T(n) = O(h) 16

17 Red-Black Trees Red-black trees are one of many balanced search tree data structures For n nodes a red-black tree has height at most (lg n + 1) = O(lg n) Assigns each node a color (red or black) Red-black properties Every node is either red or black The root node is black Every leaf (IL) is black If a node is red, then both its children are black For each node, all paths from the node to descendant leaves contain the same number of black nodes Insert and delete operations must be modified in order to maintain redblack properties (maximum height, balance condition) Use rotations to modify structure of tree while maintaining binary search tree property 17

18 Rotations x Left-Rotate(T,x) y a b y g Right-Rotate(T,y) a x b g 18

19 Insertion rbinsert(x) bstinsert(x) x.colour = red while x!= root and x.p.colour == red if x.p == x.p.p.left //parent is left y = x.p.p.right //uncle of x if y.colour == red //same as x.p x.p.colour = black y.colour = black x.p.p = red x = x.p.p else //y.colour == black if x == x.p.right x = x.p left_rotate(x) x.p.colour = black x.p.p.colour = red right_rotate(x.p.p) else //symmetric to if end while root.colour = black 19

20 Deletion rbdelete(z) if z.left == null or z.right == null y = z else y = successor(z) if y.left!= null x = y.left else x = y.right x.p = y.p //detach x from y if y.p == null //y is the root root = x else // Atttach x to y s parent if y == y.p.left //left child y.p.left = x else y.p.right = x if y!= z //i.e. y has been moved up z.data = y.data //replace z with y if y.colour = black rbdeletefixup(x) //x could be null 0

21 Tree Fixup rbdeletefixup(x) while x!= root and x.colour = black if x == x.p.left //x is left child y = x.p.right //x s sibling if y.colour == red y.colour = black x.p.colour = red //p was black left_rotate(x.p) y = x.p.right if y.l.col == blk and y.r.col = blk y.colour = red x = x.p //and into while again else if y.right.colour = black y.left.colour = black y.colour = red right_rotate(y) y = x.p.right y.colour = x.p.colour x.p.colour = black y.right.colour = black left_rotate(x.p) x = root else //symmetric to if x.colour = black 1

22 Continued Let y be the node replacing deleted node z x be the left child of y If y is red then no problem: Black-height not changed o red nodes are adjacent Root remains black

23 y is a Black ode If y is black then we may have problems: If y had been the root and a red child of y becomes the new root x and p[y] are both red Removing y causes a path to have one less black node 3