Find specific node height in tree c++ - c++

Question

I am trying traverse over tree and check for specific (rel_name) and return his height, but my function traverse over "mother" branch and checks only fathers branch.
The result is that my program return exception() and core dumping.
how do I fix my function to not core dump and check mothers branch too?
string treeHeight(Person* root, string rel_name, int height){
height++;
if(root == nullptr) {
throw exception();
}
else if(root->name == rel_name) return to_string(height);
return treeHeight(root->father, rel_name, height);
return treeHeight(root->mother, rel_name, height);
}

Answer 1

Someone has already pointed out, but if your code reaches a leaf and goes a little bit further, according to your base case it would throw an exception and exit the program.
I revised your code a little bit:
int treeHeight(Person* root, const string& rel_name){
if(root == nullptr) {
return -1;
}
else if(root->name == rel_name) return 0;
int leftHeight = treeHeight(root->father, rel_name);
int rightHeight = treeHeight(root->mother, rel_name);
if (leftHeight == -1 && rightHeight == -1) {
return -1;
} else {
return (leftHeight > rightHeight ? leftHeight : rightHeight) + 1;
}
}
Hope it helps.

Answer 2

In your code , you are not allowing the code to go the mother node, its directly from the father node call and thus its not getting executed ahead.
Try this below :
string treeHeight(Person* root, string rel_name, int height){
height++;
if(root == nullptr) {
throw exception();
}
else if(root->name == rel_name) return to_string(height);
int person_height = treeHeight(root->father, rel_name, height);
if(person_height!=0) return height; ---------> You need to apply this check
return treeHeight(root->mother, rel_name, height);
}