countNodeAtEachLevel(node * T, int a[],int level)
{
if(T!=NULL){
a[level]+=1;
countNodeAtEachLevel(T->left,a,level+1);
countNodeAtEachLeve(T->right,a,level+1);
}
}
The function is called with initial values
countNodeAtEachLevel(Root,a,0);
the array a[] gives the number of nodes at each level
a[i] denotes number of nodes at level i, where i ranges between log(n) and n
On Sun, Aug 21, 2011 at 9:47 AM, Amol Sharma <[email protected]> wrote:
> yes....bfs will do !!
> --
>
>
> Amol Sharma
> Third Year Student
> Computer Science and Engineering
> MNNIT Allahabad
> <http://gplus.to/amolsharma99>
> <http://twitter.com/amolsharma99><http://in.linkedin.com/pub/amol-sharma/21/79b/507><http://youtube.com/amolsharma99>
>
>
>
>
>
> On Sun, Aug 21, 2011 at 9:37 AM, Naman Mahor <[email protected]>wrote:
>
>> ght modification. so in bfs we use queue we have to add a dummy node in
>> queue. So this node will tell us that end of level has come.so we can find
>> out the no of node at any lev
>
>
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