Advanced Tree Structures

Transcription

1 Data Structure hapter 13 dvanced Tree Structures Dr. Patrick han School of omputer Science and Engineering South hina Universit of Technolog utline VL Tree (h 13..1) Interval Heap

2 ST Recall, inar Search Tree (ST)is a special case of inar Tree ll elements stored in the left subtreeof a node with value K have values < K ll elements stored in the right subtreeof a node with value K have values >= K alanced ST 40 Problem of ST? Unbalance ctuall affects the operating time We need a ST which can be self-balanced

3 alanced ST: VL Tree n VL (delson-velskii and Landis) Tree is a binar search tree with a balanced condition alanced condition is relaed a little bit Ever node in the tree, the height of the left and right subtreesdiffer b at most 1 alanced ST: VL Tree Node 40 violates the VL propert VL tree 6

4 VL Tree 7 alanced ST VL Tree: Insertion asicall follows insertion strateg of ST Ma cause violation of VL tree propert Rebalance is needed For Eample fter adding, 4, 6 The Tree becomes unbalance Height = 0 Height = Rebalance is needed 4 6 8

5 alanced ST VL Tree: Insertion nl nodes that are on the path from the insertion point to the root might have their balance altered Rebalance the tree at the deepest such node guarantees that the entire tree satisfies the VL propert The rebalance method is called Rotation alanced ST VL Tree: Insertion Four situations of violation: Subtrees: z 1 z Insertion happens in z 1 or z 9 h h h+ Right-Right z 1 z z 1 z Left-Left Right-Left z 1 z z 1 z Left-Right

6 alanced ST VL Tree: Insertion k z 1 z z 1 z z 1 z Right-Right z z 1 Left-Left Right-Left Left-Right RL 1 RR LL LR LR RL 17 LR 11 alanced ST VL Tree: Insertion Right-Right Left-Left Right-Left Left-Right 1

7 alanced ST: VL Tree: Insertion Rotation Two Rotations: Single Rotation Handle Left-Left and Right-Right situation Double Rotation Handle Left-Right and Left-Right situation 13 alanced ST: VL Tree: Insertion Single Rotation Right-Right z 1 z z 1 z

8 alanced ST: VL Tree: Insertion Single Rotation z 1 z 3 3 Left-Left 3 8 z 1 z alanced ST: VL Tree: Insertion Single Rotation How about Right-Left or Left-Right? Single Rotationcannot solve Right-Left and Left-Right cases Right-Left z 1 z z 1 z Left-Right 16

9 alanced ST: VL Tree: Insertion Double Rotation z 1 z z 1 z Right-Left z 1 z 17 alanced ST: VL Tree: Insertion Double Rotation Right-Left z 1 z z 1 z z 1 z

10 alanced ST: VL Tree: Insertion Double Rotation Left-Right z 1 z z 1 z z 1 z alanced ST VL Tree: Insertion Right-Right Left-Left Right-Left Left-Right Single Rotation Single Rotation Double Rotation Double Rotation 0

11 alanced ST: VL Tree: Insertion Rotation lgorithm Insert the nodeas the same as ST insertion Trace the path from the inserted node towards the root For each node encountered, check if heights of left and right subtree differ b at most 1 If es, go to net node If no, perform an appropriate rotation Stops When rotation is performed r, we ve checked all nodes in the path 1 alanced ST: VL Tree: Insertion Rotation Insert,, 3, 1, 19, Single Single Double

12 alanced ST: VL Tree: Insertion ompleit ompleit for VL Insertion is Ο(log n) 1. Insert a value to the tree: Ο(log n) Same as ordinar ST. Trace the path from the new leaf towards the root, for each node on the path: Ο(log n) heck height difference: Ο(1) If satisfies VL propert, proceed to net node If not, perform a rotation: Ο(1) hecking stops When a single rotation is performed r, we ve checked all nodes in the path 3 alanced ST VL Tree: Deletion lgorithm Delete the nodeas the same as ST deletion Trace the path from the last node deleted towards the root For each node encountered, check if heights of left and right subtree differ b at most 1 If es, go to net node If no, perform an appropriate rotation ontinue to trace the path until we reach the root Unlike insertion, more than one node ma need rotation 4

13 alanced ST: VL Tree: Deletion Single Rotation h Delete in X I h+1 h-1 X I h+1 h+1 X I h h I h-1 * The height of Inside subtree (I) can be h-1 alanced ST: VL Tree: Deletion Double Rotation I h+1 h h+1 h-1 I 1 I Delete in X I 1 I X h I 1 I X h I X I 1 h I I 1 I h-1 6

14 alanced ST VL Tree: Deletion Single Rotation h-1 h+1 k I I Double Rotation h-1 h+1 I I 1 I I 1 I 1 I 7 alanced ST VL Tree: Deletion Single Rotation I k I Double Rotation k I I 1 I I 1 I 1 I 8

15 alanced ST VL Tree: Deletion 1 6 Single Rotation Single Rotation 9 alanced ST VL Tree: Deletion 1 6 Single Rotation Double Rotation

16 alanced ST VL Tree: Deletion Delete 46, 9, 6, 9, in this VL Tree alanced ST VL Tree: Deletion Delete 46, 9, 6, 9, Single Double 3

17 Interval Heap Locate both Ma and Min values ombine with Ma- and Min-heap omplete binar tree Each node Must has elements ecept the last one [a, b] is the interval, a b The interval is contained in its parent s interval 0,70,60 30,0 1,80 18,90 1,0 46,60 30,60 3,0 Left values define a min heap Right values define a ma heap 33 Interval Heap Insert dd the value to the last node (dd new node if needed). If the node alread has a value, compare new value is bigger, the right value new value is smaller, the left value 3. Repeat to swap if the new value is bigger than the upper bound of its parent (Ma-heap) 4. Repeat to swap if the new value is smaller than the lower bound of its parent (Min-heap),90 1,80 30,60 0,70 1,0 46,60 3,0,60 30,0 18,19 34

18 Interval Heap Insert dd the value to the last node (dd new node if needed). If the node alread has a value, compare new value is bigger, the right value new value is smaller, the left value 3. Repeat to swap if the new value is bigger than the upper bound of its parent (Ma-heap) 4. Repeat to swap if the new value is smaller than the lower bound of its parent (Min-heap),90 1,80 30,60 0,70 1,0 46,60 3,0,60 30,0 16, Interval Heap Insert 8 1. dd the value to the last node (dd new node if needed). If the node alread has a value, compare new value is bigger, the right value new value is smaller, the left value 3. Repeat to swap if the new value is bigger than the upper bound of its parent (Ma-heap) 4. Repeat to swap if the new value is smaller than the lower bound of its parent (Min-heap),90 1,80 1,8 30,60 0,70 1,0 1,8 18,80 46,60 3,0,60 30,0 18,8 18,0 36

19 Interval Heap Insert 1. dd the value to the last node (dd new node if needed). If the node alread has a value, compare new value is bigger, the right value new value is smaller, the left value 3. Repeat to swap if the new value is bigger than the upper bound of its parent (Ma-heap) 4. Repeat to swap if the new value is smaller than the lower bound of its parent (Min-heap),90,90,80 1,80 30,60 0,70 1,0 46,60 3,0,60 30,0 1, Interval Heap Insert 1. dd the value to the last node (dd new node if needed). If the node alread has a value, compare new value is bigger, the right value new value is smaller, the left value 3. Repeat to swap if the new value is bigger than the upper bound of its parent (Ma-heap) 4. Repeat to swap if the new value is smaller than the lower bound of its parent (Min-heap),90 1,80 30,60 0,70 1,0 1, 46,60 3,0,60 30,0 18,

20 Interval Heap Remove Min 1. Replace the left value of root with the left value of the last node. Shift down with the lower bound if necessar 3. Swap two values if lower bound > upper bound,90 18,90 1,90 18,80 1,80 30,60 0,70 18,0 1,0 46,60 3,0,60 30,0 18, Interval Heap Remove Ma 1. Replace the left value of root with the left value of the last node. Shift down with the upper bound if necessar 3. Swap two values if lower bound > upper bound,90,19,80 1,80 1,19 1,70 30,60 0,70 0,19 19,0 19,60 1,0 46,60 3,0,60,0 0, 30,

21 Heap Time ompleit Time ompleit Heap Interval Heap Insert Remove (log(n)) (log(n)) (log(n)) (log(n)) 41