Maximum path sum from any node

/*
for each node there can be four ways that the max path goes through the node:
1)Node only
2)Max path through left child+node
3)max path  through right child + node
4)max path through left child+node+max path through right child
Note: updating max without including the root with only n1 and n2 is not the current node responsibility,it would have
been already done on the descendent node.
*/
int fun(Node *root,int &ans){
    if(root==NULL) return 0;
    int t1=0,t2=0;
    if(root->left){
        t1=fun(root->left,ans);
    }
    if(root->right) t2=fun(root->right,ans);
    int path_with_one_child=max(max(t1,t2)+root->data,root->data);  //first threee condition will be checked here
    int max_path_with_root=max(path_with_one_child,t1+t2+root->data);  //max path possible with this node as top most root of path is found here ie. 4th condition
    ans=max(ans,max_path_with_root);   //max path updated here
    return path_with_one_child;  //we are passing path generated only through condition 1,2 and 3 because only that can help to create some other max path
}                               //considering other node as top most root node

class Solution {
  public:
    //Function to return maximum path sum from any node in a tree.
    int findMaxSum(Node* root){
       int ans=INT_MIN;
          int temp=fun(root,ans);
       return ans;
    }
};