neeraj.goyal90
BAN USERUse a inverted inorder tranversal.that is go to right child first instead of left child. recursively this can be obtained as follows: The most important issue that a global count must be used when considering recursive solution.
where n=2;
void NthMax(Node *root, int n){
if(root == NULL){
return;
}
max(root->right, n);
if(--n == 0){
cout<<"Number is: "<<root->value<<endl;
}
max(root->left, n);
}
}
}
#include<stdio.h>
main()
{
int a[20][20],i,j,n,m,p,q,k=0;
printf("\n Enter Order Of matrix");
scanf("%d%d",&m,&n);
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
scanf("%d",&a[i][j]);
p=m;
q=n;
i=j=1;
while(k<p*q)
{
for(;j<=n&&k<p*q;j++)
{
printf("%d ",a[i][j]);
k++;
}
j--;
i++;
for(;i<=m && k<p*q;i++)
{
printf("%d ",a[i][j]);
k++;
}
i--;
j--;
for(;j>=i-m+1 && k<p*q;j--)
{
printf("%d ",a[i][j]);
k++;
}
j++;
i--;
for(;i>1 && k<p*q;i--)
{
printf("%d ",a[i][j]);
k++;
}
if(k<p*q)
{
j++;
i++;
n--;
m--;
}
}
}
int pow(int b, int n)
- neeraj.goyal90 November 06, 2011{
if (n == 0)
return 1;
else if (n % 2 == 0)
return pow(b, n / 2) * pow(b, n / 2);
else
return b * pow(b, n - 1)
}