5 is possible.
Considering root of the tree to be at level 0,
level 1 and level 2 are completely filled.
There are 5 internal nodes in level 1, (since all level 2 nodes are
present)
Now only (10 - 5(from level 1)+1(the root)) nodes are required.
So choose 4 nodes from level 2 and make them
I tried solving this problem and came to conclusion that none of the
options can be correct. Indeed for value 5, you can have a solution.
Then i googled for this question, and found out that, you have an
option missing. i.e 5 :|
On Aug 11, 9:01 pm, rajeev bharshetty rajeevr...@gmail.com wrote:
@amit: Thanks
On Thu, Aug 11, 2011 at 10:04 PM, amit karmakar
amit.codenam...@gmail.comwrote:
I tried solving this problem and came to conclusion that none of the
options can be correct. Indeed for value 5, you can have a solution.
Then i googled for this question, and found out that, you
* correction
(5*5(There are 5*5 nodes in level 2)-4(These became internal nodes..))
On Aug 11, 9:58 pm, amit karmakar amit.codenam...@gmail.com wrote:
5 is possible.
Considering root of the tree to be at level 0,
level 1 and level 2 are completely filled.
There are 5 internal nodes in level
Answer is 5.
the relation is
No of leaf nodes = (n-1)*(no of internal nodes) + 1
Paul
On Thu, Aug 11, 2011 at 10:44 PM, amit karmakar
amit.codenam...@gmail.comwrote:
* correction
(5*5(There are 5*5 nodes in level 2)-4(These became internal nodes..))
On Aug 11, 9:58 pm, amit karmakar
How is this formula obtained?
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I am also getting , 5 as answer now. here is what I did.
In any n-ary tree we can add one internal node by adding n-children to any
one of its leaf nodes. This operation creates one internal node and at the
cost one leaf node and adds n new leaf nodes.
L(i) be leaf nodes in a tree with i internal
