@vishal : as discussed in previous post , your solution wont work for
certain test cases...and i dont think so , checking tree in inorder way is
complex .It is simple to implement , you just need to call tree recursively
in Inorder way and keep track of prev node and compare prev node with
main()
{
int k = 5;
if (++k 5 k++/5 || ++k = 8);
printf(%d , k);
}
the output shud be 8 but it comes out to be 7.why???
as increment operator has higher precedence among them so increment shud be
done throughout at first and after then other operators shud be
I totally agree with atul007.
And that's optimal because one must check every node for
checking whether the tree is a BST or not, and this algorithm
visits each node exactly once.
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how many ways are there to add or subtract two numbers with out addition or
subtraction operator???
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yes ofcourse... dats the easiest i suppose...
but in one of my interviews, i told this approach, but was then asked not
to use space (which i was ,to store inorder)
So for such cases, you must try other approaches as well. (DO inorder,keep
track of previously visited node and compare it with
yes, that was an implementation mistake but what I meant to say was- Adding
extra check of indirect xor'ing could have a pitfall too.
Try the case: [0 1 1 1 4 4]
http://ideone.com/3sreLZ
On 4 November 2012 10:13, Vikram Pradhan vpradha...@gmail.com wrote:
It should have caught in the first
if (++k 5 k++/5 || ++k = 8); is same as if ++k 5)
(k++/5)) || (++k = 8)));
(++k 5) (k++/5) in this expression ++k5 returns false(0) ,so there is
no need to evaluate right hand side(why?)
2012/11/6 Anil Sharma anilsharmau...@gmail.com
main()
{
int k = 5;
if (++k
Hi anil
first of all when the program starts k = 5
then compiler enters if region
then k is incremented before it is compared with 5 ie k = 6
then condition is checked ie ++k 5 and it comes out to be false
then as it was end operation compiler does not evaluate the second
statement ie k++/5
then
@vaibhav : by not using extra space...i guess you mean that you were not
allowed to use one extra pointer.bcozz space complexity will remain
constant for inorder approch.
On Tue, Nov 6, 2012 at 1:07 AM, vaibhav shukla vaibhav200...@gmail.comwrote:
yes ofcourse... dats the easiest i suppose...
