How many different binary trees and binary search trees can
be made from three nodes that contain the key values 1, 2 & 3?
Answers were Sorted based on User's Feedback
Answer / hazrat hussain
Formula For BST is
{(2n)!/(n!*n)}/(n+1)
So for
N=1 BST=1
N=2 BST=2
N=3 BST=5
N=4 BST=14
N=5 BST=42
and so on
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / anonymous
30 binary trees 10 for each 1,2 and 3.
only 1 binary search tree as with 2 as root node 1 as right child and 3 as left child.
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / rohit
binary tree = n!(2^n - n).
for n =3 its 30.
for BST its 2n
C * [1/(n+1)]
n
for n=3 its 5.
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / mann
binary seach tree will be one withh 2as root node
and 1 as left child and 3 as right child of root(2).
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / mousumi
binary search tree = 5
1 1 2 3 3
\ \ / \ / /
2 3 1 3 1 2
\ / \ /
3 2 2 1
binary tees = (2^3)-3=5
| Is This Answer Correct ? | 4 Yes | 6 No |
Answer / moru laxmi narayana
18 binary trees,5 binary search trees
| Is This Answer Correct ? | 1 Yes | 3 No |
Answer / gurwinder
no.of binary trees=n!*(2^(n)-n)
no.of bst's=n*[2^(n)-n]
if 1,2,3 are key nodes
then:
no. of BST=15
no. of binary trees= 30
| Is This Answer Correct ? | 2 Yes | 4 No |
Answer / shalini
binary search tree=(2^n)-n i.e (2^3)-3=5
binary tree=((2^n)-n)*n! i.e ((2^3)-3)*3!=30
this is the right answer!!!!
| Is This Answer Correct ? | 2 Yes | 4 No |
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