Как определить требуемый размер массива дерева сегментов?
Дерево сегментов - это полное двоичное дерево. Но мы представляем его в виде массива. Помните, что для представления любого двоичного дерева высоты h в виде массива потребуется пространство, эквивалентное идеальному двоичному дереву высоты h.
[ Maximum possible Height of a Binary Tree with n nodes] (h) = Ceil[ Log_2 (n+1) ] - 1
[ No. of nodes in a Perfect Binary Tree of height h] (n) = 2 ^ (h+1) - 1
Данный массив будет представлять собой листья дерева сегментов. Так. размер данного массива будет no. листьев.
В дереве сегментов каждая пара листьев будет соединена их родителем на предыдущем уровне. И к этим родителям снова присоединятся их родители на предыдущем уровне. Так продолжается до рута.
Пример:
* Say, if there are 4 leaves in a Binary Tree, then the maximum no. of interior nodes in the Binary Tree will be N-1. So, 3.
- Then the total number of nodes in the Binary Tree = No. of interior nodes + No. of leaves. So, 4+3 = 7.
- The max possible height of this Binary Tree will be 2. Formula: Maximum possible Height of a Binary Tree (h) = Ceil[ Log_2 (n+1) ] - 1 .
- Remember, the total space required in the Segment Tree Array will be nothing but the total no. of nodes of the Perfect Binary Tree at this height.
- So, the total no. of nodes of the Perfect Binary Tree at this height is (n) = 7. Formula: No. of nodes in a Perfect Binary Tree (n) = 2 ^ (h+1) - 1.
- Thus the Segment Tree Array should also be of the size 7.
* But if there is one more leaf, say 5 and remember that this leaf can be anywhere between the beginning of the level till the end of the level.
- Then the total number of nodes in the Binary Tree = No. of interior nodes + No. of leaves. So, 5+4 = 9.
- The max possible height of this Binary Tree will be 3. Maximum possible Height of a Binary Tree (h) = Ceil[ Log_2 (n+1) ] - 1 .
- Remember, the total space required in the Segment Tree Array will be nothing but the total no. of nodes of the Perfect Binary Tree at this height.
- So, the total no. of nodes of the Perfect Binary Tree at this height is (n) = 15. Formula: No. of nodes in a Perfect Binary Tree (n) = 2 ^ (h+1) - 1.
- Thus the Segment Tree Array should also be of the size 15.
Говоря в целом,
* Say, if there are N leaves in a Binary Tree, then the maximum no. of interior nodes in the Binary Tree will be N-1.
- Then the total number of nodes in the Binary Tree = No. of interior nodes + No. of leaves. So, 2N-1.
- The max possible height of this Binary Tree will be Ceil[ Log_2 (2N) ] - 1. Formula: Maximum possible Height of a Binary Tree (h) = Ceil[ Log_2 (n+1) ] - 1 .
- Remember, the total space required in the Segment Tree Array will be nothing but the total no. of nodes of the Perfect Binary Tree at this height.
- So, the total no. of nodes of the Perfect Binary Tree at this height is (n) = 2 ^ (Ceil[ Log_2 (2N) ] ) - 1. Formula: No. of nodes in a Perfect Binary Tree (n) = 2 ^ (h+1) - 1.
- Thus the Segment Tree Array should also be of the size 2 ^ (Ceil[ Log_2 (2N) ] ) - 1.
- This can also be written as [2*2 ^ (Ceil[ Log_2 (N) ] )] - 1.
Таким образом, Размер массива дерева сегментов = [2 * 2 ^ (Ceil [Log_2 (N)])] - 1
Размер массива дерева сегментов просто 4N (приблизительно).
Пример:
Best Case Scenario: (No. of leaves (N) is a power of 2)
Say, the no. of leaves , N is 4.
Since N is a power of 2, the Segment tree will be a Perfect Binary Tree.
So the total no of nodes will be N+N-1 = 2N-1 = 7
So, the size of the Segment Tree Array = 7.
Not the Best Case Scenario: (No. of leaves (N) is not a power of 2)
If the no. of leaves , N is 5.
Since N is not a power of 2, the Segment Tree will need one more entire level to accommodate the extra 1 leaf, as this leaf can be anywhere from the beginning of the level till the end.
We know that in a Perfect binary tree, the no of nodes in every new level, will be equal to No. of all the previous level nodes + 1.
Now, total no. of nodes in the segment tree upto the previous power of 2. i.e. 8 is 8+7 = 15
So, the no. of nodes in the new level will be 15+1 = 16
So, the size of the Segment Tree Array = 15 + 16 = 31.
Говоря в целом,
Best Case Scenario: (No. of leaves (N) is a power of 2)
Since N is a power of 2, the Segment tree will be a Perfect Binary Tree.
So the total no of nodes will be N+N-1 = 2N-1
So, the size of the Segment Tree Array = 2N-1
Not the Best Case Scenario: (No. of leaves (N) is not a power of 2)
Since N is not a power of 2, the Segment Tree will need one more entire level to accommodate the extra leaves, as this leaf can be anywhere from the beginning of the level till the end.
We know that in a Perfect binary tree, the no of nodes in every new level, will be equal to No. of all the previous level nodes + 1.
Now, total no. of nodes in the segment tree upto the previous power of 2 will be 2N-1.
So, the no. of nodes in the new level will be 2N-1+1 = 2N
So, the size of the Segment Tree Array = 2N + 2N - 1 = 4N - 1 = 4N (approx.)
Таким образом, Размер массива дерева сегментов = 4N (приблизительно)
person
Vishnu Vivek
schedule
15.12.2020