The number of possible ordered trees with three nodes A,B,C is?
Answers
How to solve it

Log in to add a comment
since it is about all possibilities having all three nodes there are 12 such ordered tree 6 are binary and 6 are skewed trees

Log in to add a comment
(2ncn/(n+1)) * n! n =no of nodes so substitute n=3 answer is 30 .But there is no option 30

Log in to add a comment
This is simple. At 1st split the problem into two parts  A) Tree with 3 nodes i.e root node has two children & B) Tree with 2 nodes i.e root has one child. Now consider case 'A'. Here look root node is fixed and remaining two child nodes are able to interchange their positions. As example Let A is root node and B,C are child nodes. Now B can be left child & C can be right child. Alternatively B can be right and C can be left. So they can interchange their positions in 2! ways. Similarly If we consider B as root then we get another 2! and in case of C as root got another 2!. So from case 'A' we got total 2!+2!+2! = 6 number of distinct ordered trees. Now lets go to case 'B'. Here it's all about root and exactly one child i.e just 2 nodes. So here the node combinations are AB,BC,CA & like the case 'A' for AB we got 2!, for BC: 2! & for CA: 2!. So in case 'B' also total 2!+2!+2! = 6 number of distinct ordered trees. Now we have come to the final part. If you look carefully, then you find that case 'A' & 'B' are nonoverlapping cases as we consider distinct suit of conditions for each case respectively. So now simply add the results we got from case 'A' & 'B' i.e 6+6 = 12. Therefore, we can conclude now by our final result as 12. :) :)

Log in to add a comment
HOW TO SOLVE IT

Log in to add a comment
Related Questions
 [A] X
 [B] null
 [C] s
 [D] none of these
 [A] Optimal binary search tree construction can be performed efficiently using dynamic programming.
 [B] Breath first search cannot be used to find converted components of a graph.
 [C] Given the prefix and post fix walks over a binary tree.The binary tree cannot be uniquely constructe
 [D] Depth first search can be used to find connected components of a graph.
 [A] mn
 [B] max(m,n)
 [C] min(m,n)
 [D] m+n1
 [A] 2 deletions and 3 additions
 [B] 3 additions and 2 deletions
 [C] 3 deletions and 3 additions
 [D] 3 deletions and 4 additions