initialize array with 0 and provide appropriate value to c otherwise
it will be negative and
here c is using as index for array.
procedure visit(node root,int s[],int c) {
if (root= null)
return 0;
S[c] += n.value;
visit(root->left, c - 1)
visit(root->right, c + 1)
}
On Fri,
procedure visit(node root,int c) {
if (root= null)
return 0;
S[c] += n.value;
visit(root->left, c - 1)
visit(root->right, c + 1)
}
On Fri, Oct 21, 2011 at 1:22 PM, Mohak Naik wrote:
> If we want to compute the sum of every column, we can use the following
> algorithm
>
> a
If we want to compute the sum of every column, we can use the following
algorithm
procedure visit(node n, int c) {
if (n.left != null)
S{c} += n.value;
visit(n.left, c - 1)
visit(n.right, c + 1)
}
and call once visit(A, 0), where A is the root node. Note that S{...} in the
algorithm
You should use a doubly linked list instead of any sort of array. All the
operation on the data structure you need are goto next/prev and insert
front/end.
Yunqiao
On Wed, Oct 19, 2011 at 6:40 AM, monish001 wrote:
> I think it might done using function of following prototype:
> void func(node*
On Tue, Oct 18, 2011 at 6:31 PM, rahul patil
wrote:
> Use recursion with the code as follows
>
>
sum array of sums, with horizontal levels, level 0 is for leftmost element
>
> void add(node *root, int *sum, int level)
> {
> if(root->left)
>
{
if(level !=
Use recursion with the code as follows
void add(node *root, int *sum)
{
}
On Tue, Oct 18, 2011 at 8:58 AM, sravanreddy001 wrote:
> I that is what is suggested in the above code. Even here you can avoid the
> sorting and add all with same x coordinate.. O(n) space as suggested..
>
> --
> You
