在树 c++ 中查找特定节点高度

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

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);
}

Answer 2

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.