Data Structure - Advanced Topics in Tree -

Transcription

1 Data Structure - Advanced Topics in Tree - AVL, Red-Black, B-tree Hanyang University Jong-Il Park

2 AVL TREE Division of Computer Science and Engineering, Hanyang University

3 Balanced binary trees Non-random insertion to BST can produce unbalanced trees Can we rebalance the tree so that the tree always has O(log n) height? We need balance information for each node. Balancing the root is not enough Balancing every node is too strict Division of Computer Science and Engineering, Hanyang University

4 AVL tree AVL tree is a binary search tree. Def: For every node in the tree, the heights of its left subtree and right subtree differ by at most 1. (the height of a null subtree is 1) Is this condition enough to guarantee the height of an AVL tree with n nodes is O(log n)? Division 4 of Computer Science and Engineering, Hanyang University

5 AVL tree with Minimum Number of Nodes N 0 = 1 N 1 = 2 N 2 = 4 N 3 = N 1 + N = 7 height of left=? height of right=? Division of Computer Science and Engineering, Hanyang University

6 Smallest AVL tree of height 9 Smallest AVL tree of height 7 Smallest AVL tree of height 8 Smallest AVL tree of height 9 Division of Computer Science and Engineering, Hanyang University

7 Height of AVL tree Denote N h the minimum number of nodes in an AVL tree of height h N 0 = 1, N 1 =2 (base) N h = N h-1 + N h-2 +1 (recursive definition) N > N h = N h-1 + N h-2 +1 > 2 N h-2 > 4 N h-4 > > 2 i N h-2i If h is even, let i = h/2 1. The equation becomes N > 2 h/2-1 N 2 N > 2 h/2-1 4 h = O(log N) Ifh is odd, let i = (h 1)/2. The equation becomes N > 2 (h-1)/2 N 1 N > 2 (h-1)/2 2 h = O(log N) Thus, many operations (i.e. searching) on an AVL tree will take O(lo g N) time 7 Division of Computer Science and Engineering, Hanyang University

8 Insertion into AVL tree Insertion routine is the same as BST, but if the resulting tree is unbalanced, we should rebalance it through rotating start from the node inserted, travel up the tree, and update balance info. if there s a bad node, rotate to resume balance Rotating is a local operation and takes a constant amount of time Division 8 of Computer Science and Engineering, Hanyang University

9 Rotation If the root of T is unbalanced after insertion of x, Single Rotation: if x < T->Left->Element (Rotate_Left) [LL] or if x > T->Right->Element (Rotate_Right) [RR] k2 k1 k1 k2 x y z x y z Division 9 of Computer Science and Engineering, Hanyang University

10 Rotation If the root of T is unbalanced after insertion of y, Does Single Rotation work? k2 k1 k1 k2 x y z x y z Division 10 of Computer Science and Engineering, Hanyang University

11 Rotation Double Rotation: if x > T->Left->Element (Rotate_Left) if x < T->right->Element (Rotate_Right) [LR] or [RL] k3 k2 w k1 x k2 y z k1 k3 w x y z k3 k2 w k2 k1 z k3 k1 w x y z x y Division 11 of Computer Science and Engineering, Hanyang University

12 From Wikipedia Division of Computer Science and Engineering, Hanyang University

13 Homework: Rotation exercises Single rotation Insert sequence: 3, 2, 1, 4, 5, 6, 7 Double rotation Insert sequence: 4, 2, 6, 1, 3, 5, 7, 16, 15, 14, 13, 12, 11 Another exercise Insert sequence: 2, 1, 4, 5, 9, 3, 6, 7 Insert sequence: 3, 2, 1, 4, 5, 6, 7, 16, 15, 14, 13, 12, 11, 10, 8, 9 Division 13 of Computer Science and Engineering, Hanyang University

14 Code for Rotations struct AvlNode; typedef struct AvlNode *Position; typedef struct AvlNode *AvlTree; struct AvlNode { ElementType Element; AvlTree Left; AvlTree Right; int Height; } static int Height(Position P) { if (P == NULL) return 1; else return P->Height; } Division 14 of Computer Science and Engineering, Hanyang University

15 Code for Rotations SingleRotateWithLeft( Position K2 ) { Position K1; k2 k1 K1 = K2->Left; K2->Left = K1->Right; K1->Right = K2; x k1 y z x y k2 z K2->Height = Max( Height( K2->Left ), Height( K2->Right ) ) + 1; K1->Height = Max( Height( K1->Left ), K2->Height ) + 1; return K1; /* New root */ } DoubleRotateWithLeft( Position K3 ) { K3->Left = SingleRotateWithRight( K3->Left ); return SingleRotateWithLeft( K3 ); } k3 k2 k1 k1 z k3 k2 w w x y z x y Division 15 of Computer Science and Engineering, Hanyang University

16 Code for insertion Insert( ElementType X, AvlTree T ) else if( X > T->Element ) { { T->Right = Insert( X, T->Right ); if( T == NULL ) { if( Height( T->Right ) Height( T->Left ) == 2 ) T = malloc( sizeof( struct AvlNode ) ); if( X > T->Right->Element ) if( T == NULL ) T = SingleRotateWithRight( T ); FatalError( Out of space!!! ); else else{ T = DoubleRotateWithRight( T ); T->Element = X; T->Height = 0; } T->Left = T->Right = NULL; /* Height adjustment when inserted without } rotation */ } T->Height = Max(Height( T->Left ), Height( T->Right ))+1; else if ( X < T->Element ) { return T; T->Left = Insert( X, T->Left ); } if( Height( T->Left ) Height( T->Right ) == 2 ) if( X < T->Left->Element ) T = SingleRotateWithLeft( T ); else T = DoubleRotateWithLeft( T ); } Division 16 of Computer Science and Engineering, Hanyang University

17 Deletion from an AVL tree May need O(log n) rotations when deleting a node, in the worst case. D E M A F X B G I C H J K L Division 17 of Computer Science and Engineering, Hanyang University

18 RED BLACK TREE Division of Computer Science and Engineering, Hanyang University

19 Red black tree Red black tree is a binary search tree in which every node is colored either red or black. All properties are based on the extended binary search tree; each null pointer is replaced with an external node. A pointer to a black child (including the external node) is black; a pointer to a red child is red. Properties of colored nodes root and all external nodes are black no consecutive red node is on the root-to-external node path all root-to-external node paths have the same number of black nodes Division 19 of Computer Science and Engineering, Hanyang University

20 Red black tree Rank (black height) of a node is the number of black pointers on any path from the node to any external node the rank of an external node is 0 Lemma 1 Let the length of a root-to-external-node path be the number of pointers on the path. If P and Q are two root-to-external-node paths in a red-black tree, length(p) 2 length(q) Proof: When r is the rank of the root, each root-to-external-node path has between r and 2r pointers Division 20 of Computer Science and Engineering, Hanyang University

21 Red black tree Division 21 of Computer Science and Engineering, Hanyang University

22 Insertion If a new node is colored in black, we will have an extra black node on paths require recoloring If a new node is colored in red, we might have two consecutive red nodes may or may not need recoloring Insert first make a normal insert into a binary search tree color it with red fix red-black properties Division of Computer Science and Engineering, Hanyang University

23 Insertion LLr new node LRr new node Division of Computer Science and Engineering, Hanyang University

24 Insertion a b LLb b c a c new node a a c LRb b c b a c new node b Division of Computer Science and Engineering, Hanyang University

25 Insertion insert 70, 60, 65, and 62 in order. Division 25 of Computer Science and Engineering, Hanyang University

26 Deletion We can delete a node with any external node in binary search tree delete a node without any child or with one child in the binary search tree When the node with two internal nodes is deleted, find the node of its predecessor or successor and delete that node delete a node with both children in the binary search tree delete 65 delete Division of Computer Science and Engineering, Hanyang University

27 Deletion If a red node is deleted, no rebalancing is needed since the rank is not changed (property #3) If a black node is deleted, rebalancing is needed color the edge double black Division of Computer Science and Engineering, Hanyang University

28 Deletion: how to remove the double edge black sibling with a red child - a node in blue can be either black or red - a node in dotted line can exist or not exist p s v s p z z v p z v s p s z v a b a b Division of Computer Science and Engineering, Hanyang University

29 Deletion: how to remove the double edge black sibling with black children (including any internal node) p p v s v s p p propagate upward! v s v s Division 29 of Computer Science and Engineering, Hanyang University

30 Deletion: how to remove the double edge red sibling restructure to have a black sibling p s s v s p b p b a b v a v a Division 30 of Computer Science and Engineering, Hanyang University

31 Homework: Deletion delete 9, 8, 7 Division 31 of Computer Science and Engineering, Hanyang University

32 B TREES Division of Computer Science and Engineering, Hanyang University

33 B-Trees Binary Trees are not quite appropriate for data stored on disks disk access is MUCH slower than memory access disk is partitioned into blocks (pages) and the access time of a word is the same as that of the entire block containing the word. We have to reduce the number of disk accesses Make each node of the tree wider (multi-way search tree) k k1 k2 k3 p0 p1 p2 p3 x < k x > k x < k1 k1<x<k2 k2<x<k3 k3 < x Division 33 of Computer Science and Engineering, Hanyang University

34 B-Trees (Rudolf Bayer 1972) Division 34 of Computer Science and Engineering, Hanyang University

35 An Example B-Tree with m = Tree 18 : 13 : 30 : 45 8 : : 22 : : 58 : 59 Division 35 of Computer Science and Engineering, Hanyang University

36 Height of a B-Tree Division 36 of Computer Science and Engineering, Hanyang University

37 Node Structure #define order 32 struct B_node { int n_child; /* number of children */ B_node *child[order]; /* children pointers */ int key[order-1]; /* keys */ } Division 37 of Computer Science and Engineering, Hanyang University

38 Search When we arrive at an internal node with key k 1 < k 2,... < k m-1, search for x in this list (either linearly or by binary search) if you found x, you are done otherwise, find the index i such that k i < x < k i+1 (k 0 = and k m = ), and recursively search the subtree pointed by p i. Complexity = log m log m n= O(log n) Division 38 of Computer Science and Engineering, Hanyang University

39 Insertion Do a search to find the appropriate leaf into which to insert the node if the leaf is not full (has < m 1keys), simply insert it if the node overflows, restore the balance (1) Key-Rotation: Check for Siblings for rotation 15 : : 42 9 : : 22 : 30 9 : : 30 T1 T1 Key-Rotation is convenient but neither sufficient nor necessary Division 39 of Computer Science and Engineering, Hanyang University

40 Insertion Do a search to find the appropriate leaf into which to insert the node if the leaf is not full (has < m 1keys), simply insert it if the node overflows, restore the balance (1) Key-Rotation: Check for Siblings for rotation 15 : : 42 9 : : 22 : 30 42: 9 : : : : 30 : 9 9 : 15 T1 18 : : 42 T1 Key-Rotation is convenient but neither sufficient nor necessary Division 40 of Computer Science and Engineering, Hanyang University

41 Insertion Division of Computer Science and Engineering, Hanyang University 41

42 8 : 19 3 : 12 : : 40 insert(36) 1 : 5 : 9 : 14 : 17 : 20 : : : 55 8 : 19 overflow! 3 : 12 : 16 1 : 5 : 9 : 14 : 17 : 8 : 19 : : 35 : : : 36 : 51 : 55 overflow! 3 : 12 : : 40 : 1 : 5 : 9 : 14 : 17 : 20 : : 36 : 51 : : 8 : 3 : 12 : : 27 : 40 : 1 : 5 : 9 : 14 : 17 : 20 : : 36 : 51 : 55 Division 42 of Computer Science and Engineering, Hanyang University

43 Deletion We still need to find a suitable replacement which is the largest key in the left child (or the smallest in the right) and move it to fill the hole. The replacement is always at the leaf level and creates a hole in a leaf node. if the leaf still has sufficient capacity, we are done. otherwise, node merging is performed. Division 43 of Computer Science and Engineering, Hanyang University

44 Merging Division of Computer Science and Engineering, Hanyang University 44

45 8 : 19 3 : 12 : 16 1 : 5 : 9 : 14 : 17 : 27 : : : : 55 delete(5) 8 : 19 3 : 12 : : 40 1 : 9 : 14 : 17 : 20 : : : 55 underflow! 8 : 19 Node Merge 12 : : 40 1 : 3 9 : 14 : 17 : 20 : : : : 19 8 : 16 : 27 : 40 Key Rotation 1 : 3 9 : 14 : 17 : 20 : : : 55 Division 45 of Computer Science and Engineering, Hanyang University

46 Difference from Weiss textbook (B+ Tree) All records are allowed to be stored only in the leaves. All non-leaf nodes include only the key values which can be found in the subtrees of the node. Leaf nodes can have a pointer to its next sibling so that a sequential access is possible Division 46 of Computer Science and Engineering, Hanyang University

47 Practical Use of B-Tree Division 47 of Computer Science and Engineering, Hanyang University