Download presentation

Presentation is loading. Please wait.

Published byBritney Franklin Modified over 3 years ago

1
One more definition: A binary tree, T, is balanced if T is empty, or if abs ( height (leftsubtree of T) - height ( right subtree of T) ) <= 1 and if the left subtree of T and the right subtree of T are balanced. That is, a binary tree, T, is balanced if for every node N of T, abs ( height (left subtree of N) - height (right subtree of N) ) <= 1

2
balanced Unbalanced at this node

3
The basic reason for using a binary search tree for storing data is to optimize search time. If a binary search tree, T, containing n nodes is full, this optimized search time is realized, and n = 2 k - 1 for some non-negative integer k The height (T) = k - 1 The tree T has k levels. The tree T has nodes at levels 0 (root node), 1, 2, 3... k - 1 (and all leaf nodes of T are at level k - 1. Solving the equation above gives k = log 2 (n + 1)

4
n = 31 n = 2 k - 1 k = log 2 (n + 1) = 5 The height of the tree is k - 1 = 4 All leaf nodes are at level 4.

5
At the opposite extreme, a binary search tree with n nodes could be a chain, or vine. A tree such as this one has n levels, and a height of n - 1 The maximum number of comparisons in a search for a particular node is n - one for each level.

6
The maximum number of comparisons in a search of a full binary search tree for a particular node is k = log 2 (n + 1) - one comparison at each level. The maximum number of comparisons in a search of a binary search tree which is a chain for a particular node is n - one for each level. For example if n = 262,143 If the n nodes are stores in a full binary search tree, a maximum of 18 comparisons are needed in a search for a particular node. If the n nodes are stored in a binary search tree that forms a chain, a maximum of 262,143 comparisons are needed in a search for a particular node.

7
Clearly a full binary search tree realizes the optimal search time. But a full tree has no room to grow, or shrink. Between these two extremes, a balanced tree or a complete tree yields close to the optimal search time, and still has room to grow.

8
The minimum height of a binary tree with n nodes is ceiling ( log 2 ( n + 1) ) - 1 And the number of levels is ceiling (log 2 ( n + 1) ) To show this: Let k be the smallest integer for which n <= 2 k - 1 Then 2 k-1 - 1 < n <= 2 k - 1 Add one to all three parts of this inequality and take the log 2 of all three parts: k - 1 < log 2 ( n + 1) <= k

9
If the equality holds, the tree is full, and k = log 2 ( n + 1) = the number of levels height of the tree = log 2 ( n + 1) - 1 Otherwise, log 2 ( n + 1) is not an integer; round it up, and k = ceiling (log 2 ( n + 1) ) = the number of levels and ceiling (log 2 ( n + 1) ) - 1 = the height of the tree.

10
Suppose T is a binary search tree with 300,000 nodes having minimal height, for instance T may be a complete tree. The smallest integer, k, for which n <= 2 k - 1where n = 300,000 Is 192 19 - 1 = 524, 287 2 18 - 1 = 262,143 So 2 k-1 - 1 < n <= 2 k - 1 And the maximum number of comparisons in a search of this binary search tree for a particular node is 19 And if T is a complete tree, there are 150,000 leaf nodes, more than 112,000 are at next lowest level so the tree can grow without degrading search times.

11
In practice searching a set of data occurs MUCH MORE frequently than adding a new item of data, or removing an existing item of data. The algorithm presented in the text follows the premise that whenever a node is added to, or removed from, a balanced tree, the tree is tested, and if is unbalanced, the tree is rebalanced with one or more rotation operations.

12
A newer algorithm that will be presented in class follows a different premise. The tree is initially built as a complete binary search tree. As nodes are added, and removed (following the algorithms illustrated in class), the tree may become closer to a chain, and further from a complete tree. Consequently the search times become degraded. A statistical utility tracks the search times, and when the average number of comparisons per search exceeds log 2 (n + 1) by some percentage, a rebalancing utility is called to reform the binary search tree to a complete binary search tree. So rebalancing occurs only when performance is suffering.

13
The rebalancing algorithm 1. Converts the tree to a vine. A vine is a binary tree in which the left child of every node is NULL. 2.Convert the chain to a complete binary search tree.

14
Step One - converting the tree to a vine: For each node, N, in the tree if N has a left child rotate N and its left child to the right (clockwise). If the left chilld of N has a right subtree, that subtree becomes the left subtree of N.

15
Step Two - converting the vine to a complete binary tree. The sequence { 2 k - 1: k >= 1} = { 1, 3, 7, 15, 31,... } plays an important role in this step. Let n = the number of nodes in the vine. Let k be the smallest integer so that 2 k-1 - 1 < n <= 2 k - 1

16
Case I: n = 2 k - 1 - the resulting complete tree will be a full tree. 1. At every second node, N, (and its parent), rotate to the left (counter clockwise). If N has a left subtree, it becomes the right subtree of N’s parent. The number of rotations = 2 k-1 - 1 ( a value in the sequence above). 2. Repeat these rotations at every second node in the right chain for 2 k-2 - 1 nodes (the next smaller value in the sequence above). At the last repetition, perform a single left rotation at the second node and its parent.

17
Case II: 2 k-1 - 1 < n < 2 k - 1 The resulting tree will be complete, but not full. 1. Do a left rotation about every second node for a total of n - (2 k-1 - 1) nodes. This is the number of nodes that will be in the lowest level of the complete tree. The resulting chain of right children will contain 2 k-1 - 1 nodes. 2. Apply Case I.

Similar presentations

OK

Topic 23 Red Black Trees "People in every direction No words exchanged No time to exchange And all the little ants are marching Red and black antennas.

Topic 23 Red Black Trees "People in every direction No words exchanged No time to exchange And all the little ants are marching Red and black antennas.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on applied operational research mathematics Free ppt on motivation for students Ppt on human chromosomes disorders Science ppt on carbon and its compounds Ppt on chromosomes and genes in dna Ppt on management by objectives method Ppt on hospitality industry in india Download ppt on oxidation and reduction examples Download ppt on home security system Ppt on council of ministers of education