树的路径中的最大总和是多少

Question

https://practice.geeksforgeeks.org/problems/maximum-path-sum/1 this is question from geeks for geeks. i wrote my ans. but it is giving wrong. what is problem with my logic?

int path(Node *root, int & max_sum)
{
  if(root==NULL)
    return 0;
  int l=path(root->left,max_sum);
  int r=path(root->right,max_sum);
  max_sum=max(max_sum,l+r+root->data);
  return max(l,r)+root->data;
}

int maxPathSum(Node *root) 
{
  int max_sum=INT_MIN;
  path(root,max_sum);
  return max_sum;
  // code here
}

Answer 1

The problem statement says

Find the maximum possible sum from one leaf node to another.

So, when you do

 max_sum=max(max_sum,l+r+root->data);

You will need to check whether the root has both the children. Otherwise, they won't be considered in the final answer .

You also did

if(root==NULL)  return 0;

and

max(l,r)+root->data;

These both combined are overriding a single leaf node child whose value is negative since 0 is greater than any negative integer.

So overall your code should look like this:

int path(Node *root, int & max_sum)
{
  if(root==NULL) return 0;
  int l=path(root->left,max_sum);
  int r=path(root->right,max_sum);
  if(root->left != NULL && root->right != NULL){
      max_sum = max(max_sum,l + r + root->data);
  }

  if(root->left != NULL && root->right == NULL) return  l + root->data;
  if(root->right != NULL && root->left == NULL) return  r + root->data;
  return max(l,r) + root->data;
}

int maxPathSum(Node *root) 
{
  int max_sum=INT_MIN;
  path(root,max_sum);
  return max_sum;
}