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The number of ways in which the numbers $1, 2, 3, 4, 5, 6, 7$ can be inserted in an empty binary search tree, such that the resulting tree has height $6$, is _________.

Note: The height of a tree with a single node is $0$.
asked in DS by Veteran (41.3k points)  
edited by | 2.8k views

7 Answers

+21 votes
Best answer

 

WELL its RECURRENCE QUESTION ,,,,but for that need to observe the pattern ..first and then THE 
RECURRENCE RELATION WILL BE N(h)=2N(h-1)

answered by Active (1.8k points)  
selected by
Thanks Deepesh superb explanation.

$N(h)= 2N(h-1)$ is the recurrence

& if we not take extreme elements then it is BST less than height 6 or it is not at all a BST.

So every level 2 Possibilities.
+41 votes
We need to fill 7 levels with 7 elements. So, at each level we have exactly 2 possible options like 1 and 7 for root- one corresponding to making it left skewed and other right skewed. And this is the same for all levels up to 6 giving $2^6 = 64$ possible ways.
answered by Veteran (281k points)  

Hi ... @Arjun Sir .. 

Please explain the difference between the above question and 

question mentioned here 

http://gateoverflow.in/3462/gate2007-it-29

+9 votes
there can be total 2^6 skew tree ,and being a BST there is only one permumation for a given BST .

ytotal 2^6 diffrnect skew trees are possible
answered by Active (2.1k points)  
+5 votes

But  , here your resulting tree height must be 6.......

so it is only possible when your tree is either left-skew or right-skew ....

and above tree is binary search tree..........

so either you can insert in increasing order (right-skew tree) or insert in decreasing order (left-skew tree).....

so only possible way is 2....................

Emphasized point=>  (1) height must be 6      (2)  Binary search tree

think about it !!!!

answered by (97 points)  
left-skew or right-skew. But his can be changed for any level- at each level we got 2 options. So, $2^6=64$ ways (we don't need the entire tree to be left or right skewed).
same vith me..answer is 2
But , @arjun sir , still confused the below condition must satisfy for above question (1) height must be 6 ( root element height is 0)... (2) tree must be binary search tree... (3) element are 1,2,3,4,5,6,7..... so , either you can insert your element in (increasing order)[1,2,3,4,5,6,7](right-skew tree) or decreasing order [7,6,5,4,3,2,1] left skew tree........ no other option are available....... if you try something other option .... tree's height not be 6............
ohk......got it arjun sir....clear the confusion.......
4

                                         7

                                   6

                            5

                   3

            2

     1

 

this is one kind of tree possible so ur answer is wrong
you can also from  start from 1 and make the same tree bending towards right and more like taking 1 as root after that seven as right child then 5 as left child and so on ..... and many more keep in mind that at each level you have to insert only single node and for each location you have two choice either choose smaller one or bigger one and their are 6 level so 2^6 = 64 possible trees
+4 votes
A simple way to solve this question:
See that we need height of 6 and we have 7 elements at hand. So obviously any node in BST can't have two children.Now think about if root node is 2 or 3 or 4 or 5 or 6,then that root must have 2 children.So root must be either 1 or 7.Let denote root as 1st level.
So at 1st level, we either can choose 1 or 7(2 choices)
At 2nd level,similarly we can choose 2 or 6(2 choices)
At 3rd level,similarly we can choose 3 or 5(2 choices)
At 4th level,similarly we can choose 4 or 4(2 choices)
At 5th level,similarly we can choose 5 or 3(2 choices)
At 6th level,similarly we can choose 6 or 2(2 choices)
At 7th level, we will have only 1 number remaining.
So total number of BSTs possible=2^6=64
answered by (291 points)  
0 votes

The Recurrence Relation for the no. of  binary search tree for 'n' nodes(distinct) with height 'n-1' will be  

NBST(n-1)=NBST(n-2)+NBST(n-2) ;

Base condition:NBST(0)=1 and NBST(1)=2 ;

So NBST(6) =2*NBST(5)=2*2*NBST(4)=2*2*2*NBST(3)=2*2*2*2*NBST(2)=2*2*2*2*2*NBST(1)=26=64.

Eg:Take 3 nodes 1,2,3.No.of BST with height 2 is (i)With root as 1 ,the no. of possible BST will be 2(ie.NBST(1)) , (ii)With root as 3 ,the no. of possible BST will be 2(ie.NBST(1)) and (iii)With root as 2,there will not be any BST of height 2.

So 2+2 =4.

answered by (381 points)  
reshown by
0 votes
Arrange first all the nodes in asc or dsc order
Now observe 2 bottom nodes
U can come up with 2 bsts with that only 2 nodes
Now with three bottom nodes u get 4 bsts
Like wise
U get 2^6 at level 0
answered by (333 points)  


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