Binary-search-trees-in-JavaScript
No. | Questions |
---|---|
Binary search tree | |
1 | Binary-search-tree-and-its-applications |
2 | Height-balanced-BST |
3 | BST Insertion |
4 | Inorder successor and predecessor |
5 | BST Delection |
| 1 | Binary-search-tree-and-its-applications
Left sub -tree :LT, Right sub- Tree: RT, Where P: Parent
Let we assume that 'LEFT' :sub-tree ≤ P' is called left side smaller
Let we assume that 'RIGHT' :sub-tree ≥ P' is called right side is bigger
same like wise LEFT PARENT value
KEY REPRESENTS values of the tree and CURRENT tells the value of PARENT
When ever KEY values LESS then the CURRENT PARENT value then its represents the LEFT side of the tree
Same like wise RIGHT PARENT value
KET REPRESENTS values of the tree and CURRENT tells the value of PARENT
When ever KEY values GREATER then the CURRENT PARENT value then its represents the RIGHT side of the tree
select the RIGHT side elements of the KEY values which are GREATER than PARENT TREE
Select the LEFT side elements of the KEY values which are LESS than PARENT TREE
KEY === CURRENT is equal to parent value
DISTENCE From root node to fargest d done is 4
Distence feom fargest d node to root node is HEIGHT of the Tree
search is taken as o(h) time,I can search any type of tree
height of the Tree is also 4 and the number of compressions is also 4
IMP 11.19
Binary Search search TREE=> o(h) and in an unsorted array |---------------| o(n)
search time complexity is o(h) times
prestrict the hight of the array in an unsorted array we need to ettrate o(n) times
let say construct a tree a binary search tree then h =log(n) the SEARCH will becomes =>o(h) means SEARCH => o(logN)
Search complexity improve dramatically o(n), if h == logn
Search=> o(logn)
Nodes , N =7 anf HEIGHT , h=log7 2
|h(l) - h(r)| ≤ 1
if h == logN NODES =7 AND HEIGHT H = log 7 2 => 3
IF h == logN NODES =4 and HEIGHT h = log 4 4 =>
in a stand binary standard there is non restrection
In a balanced binary search tree it is directly it is h= o(logn) and search = o(logn)
| 2 | Height-balanced-BST
Red Black Tree
AVL trees
'WHITE", 'RED BLACK TREE " are important when compared among these 'AVL" is more important in Binary search tree
| 3 | BSTinsertion
Let insert new element 13 and compare from root element and find correct position for 13
Attach the new node
First compare Root Node 10 is (greater then or equal) or (less then or equal) BST as a rule is that root element is greater then ket lements it is left side of the BST
Second if is some case ROOT element is less than KEY elements then it is belongs to RIGHT HAND side of the tree
INSERT '13' > 10 ------> it belongs to LEFT SIDE of the TREE
INSERT 13 < 15 ------> it belongs to RIGHT SIDE of the TREE
INSERT 13 > 12 ------> it belongs to LEFT SIDE of the TREE that means 12 RIGHT SIDE , LAST ELEMENT
INSERT 16' > 10 ------> it belongs to RIGHT SIDE of the TREE
INSERT 16 > 15 ------> it belongs to RIGHT SIDE of the TREE
INSERT 16 < 19 ------> it belongs to LEFT SIDE of the TREE that means 19 LEFT SIDE, LAST ELEMENT
TC => O(h)
HEIGHT OF THE TIME COMPLEXITY
| 4 | Inorder successor and predecessor
Inorder successor
Inorder predecessor
Inordersuccessor 15(9) = 10 15(3) = 5 15(12) = 15
Inorder predecessor Ip(9) = 8 Ip(7) = 5 Ip(12) = 11
| 5 | BTS Delection
these are dependes on nodes and key values
No child in in BST delection type