Question

this is study material, not homework:

I have the following tree, I need to write an algorithm that finds a given number and returns an integer indicating how many nodes it visited before finding it. It should also print the values of all "ancestor" nodes relative to the node in which the value was found (in no particular order, and it is assumed that the given value is always present)

If the given value is 40 it should return 4 and print 30, 20, 10 (in any order)

I've written the following solution, and I think it works, but I'm concerned about the print.

void foobar (ty_tree *tree, int value, int & count){
    if (tree !=null) {
        if (tree->value != value) {
            count++;
            foobar (tree->left, value, count);
            foobar (tree->right, value, count);
            cout << tree->value;          
        }
    }
}

Answer 1

好方法！但是要打印祖先（即父节点），如果在其中一个子节点中找到了值，则需要在递归函数中知道：

bool foobar (ty_tree *tree, int value, int & count) {
    if (tree !=nullptr) {  // oops: NULL or nullptr, the latter is better
        if (tree->value != value) {
            count++;
            if (foobar (tree->left, value, count) ||
                  foobar (tree->right, value, count) ) // if found below
            cout << tree->value<<endl;  // print the node, because it's on the path          
        }
        else {
            cout << "Found: "<<tree->value<<endl;  // print the value found
            return true;     // and inform caller that he can print as well.
        }
    } 
    else return false;  // reached a leaf without finding        
}

由于评论中表达了一些疑问，这里有online demo

Recursive tree, printing ancestor nodes

1 个答案: