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, Oct 21, 2011 at 8:27 PM, ashish shukla
<[email protected]>wrote:
> 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 <[email protected]> wrote:
>
>> If we want to compute the sum of every column, we can use the following
>> algorithm
>>
>> and call once visit(A, 0), where A is the root node. Note that S{...} in
>> the algorithm is a map whose keys are the columns numbers (..., -2, -1, 0,
>> 1, 2, ...) and whose values (at the end of the algorithm) are the sums of
>> the values of nodes in that column (S{1} will be the sum of nodes in
>> column 1). We can also use an array, instead of a map, provided that we pay
>> attention to the indexes (arrays have no negative indexes). The algorithm is
>> still O(n), because we traverse the entire tree only once. However, in this
>> case we need additional space to store the sum for all columns (the map, or
>> the array).
>>
>> ----stackoverflow solution.
>>
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>
>
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