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 mohak...@gmail.com wrote:
If we want to compute the sum of every column, we can use the following
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,
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
I think it might done using function of following prototype:
void func(node* root, dequeint d, const dequeint::iterator it);
I will add left child's value in it-1 if exists else create new...
similarly for right child.
and call the same function for each of the children to explore
further..
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 monish.gup...@gmail.com wrote:
I think it might done using function of following
Use recursion with the code as follows
void add(node *root, int *sum)
{
}
On Tue, Oct 18, 2011 at 8:58 AM, sravanreddy001 sravanreddy...@gmail.comwrote:
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
On Tue, Oct 18, 2011 at 6:31 PM, rahul patil
rahul.deshmukhpa...@gmail.comwrote:
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)
{
I have an alternate method.
At every level, maintain records with (x-axis value, nodeValue) and
keep them in a queue.
After traversal is done, you can sort the elements based on x-Axis
value and sum up the nodeValue for records having same x-axis value.
But complexity is becoming O(nlogn)
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..
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