int maxLtoLSum(NODE *root,int *sum)
{
int ls=0,rs=0,lsum=0,rsum=0;
if(root==NULL)
{
*sum=0;
return 0;
}
lsum= maxLtoLSum (root->left,&ls);
rsum= maxLtoLSum (root->right,&rs);
*sum=max(ls,rs) + root->data;
return max(ls + rs + root->data, max(lsum,rsum));
}
On Tue, Aug 28, 2012 at 4:02 PM, vaibhav shukla <[email protected]>wrote:
> I guess it might be
> finding maximum sum path for left subtree + max sum path for right subtree
> + root->data
>
> As in case of finding the diameter which say
> height(root->left)+height(root->right)+1
>
> Please correct me!
>
>
> On Mon, Aug 27, 2012 at 11:46 PM, kunal rustgi <[email protected]>wrote:
>
>> @ravi
>> yep, you're right.
>>
>> But method is similar to finding diameter as given on geeksforgeeks as
>> atul suggested . Thanks.
>>
>>
>> On Mon, Aug 27, 2012 at 11:23 PM, Ravi Maggon <[email protected]>wrote:
>>
>>> @atul:
>>> I think he is asking for max. sum of elements between 2 leaf nodes and
>>> not the max distance between two nodes.
>>>
>>>
>>> On Sun, Aug 26, 2012 at 6:12 PM, atul anand <[email protected]>wrote:
>>>
>>>> its the diameter of tree.
>>>> you can find implementation on geeksforgeeks
>>>>
>>>> On 8/25/12, kunal rustgi <[email protected]> wrote:
>>>> > Hi,
>>>> >
>>>> > Can anyone suggest the best approach for finding max sum b/w 2 leaf
>>>> nodes
>>>> > in a binary tree ( not BST ) ?
>>>> >
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>>>>
>>>
>>>
>>> --
>>>
>>> ....................................
>>>
>>> *Regards
>>> *Ravi Maggon
>>>
>>> Member Technical - IT/Front Office
>>>
>>> D.E. Shaw & Co.
>>>
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>>
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>
>
>
> --
> best wishes!!
> Vaibhav
>
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