3
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I'm writing C++ for the first time and wanted to implement a BST. I come from writing C, so I tend to write code in that way. I'm trying to wrap my head around smart pointers, but can't seem to get them to work properly. This version of the code is without any smart pointers. Any suggestions?

The C++ file:

#include <iostream>
#include "bst.h"

int main()
{
int i = 0;
BST<int> *tree = new BST<int>;

tree->addNode(new node<int>(6));
tree->addNode(new node<int>(5));
tree->addNode(new node<int>(2));
tree->addNode(new node<int>(5));
tree->addNode(new node<int>(7));
tree->addNode(new node<int>(8));

std::cout << "In order ";
tree->inOrderPrint(tree->getRoot());
std::cout << std::endl;
std::cout << "Pre order ";
tree->preOrderPrint(tree->getRoot());
std::cout << std::endl;
std::cout << "Post order ";
tree->postOrderPrint(tree->getRoot());
std::cout << std::endl;
std::cout << "Level order " << std::endl;
tree->levelOrderPrint(tree->getRoot());
std::cout << std::endl;

std::cout << "Minimum value " << tree->findMin(tree->getRoot())->getData();
std::cout << std::endl;
std::cout << "Maximum value " << tree->findMax(tree->getRoot())->getData();
std::cout << std::endl << std::endl;

std::cout << "Successor of 5: ";
if(node<int> *temp = tree->getSuccessor(tree->findNode(5)))
{
    std::cout << temp->getData() << std::endl;
} 
else
{
    std::cout << "None" << std::endl;
}

std::cout << "Successor of 8: ";
if(node<int> *temp = tree->getSuccessor(tree->findNode(8)))
{
    std::cout << temp->getData() << std::endl;
} 
else
{
    std::cout << "None" << std::endl;
} 

std::cout << "Predecessor of 7: ";
if(node<int> *temp = tree->getPredecessor(tree->findNode(7)))
{
    std::cout << temp->getData() << std::endl;
} 
else
{
    std::cout << "None" << std::endl << std::endl;
}

std::cout << "Predecessor of 2: ";
if(node<int> *temp = tree->getPredecessor(tree->findNode(2)))
{
    std::cout << temp->getData() << std::endl;
} 
else
{
    std::cout << "None" << std::endl << std::endl;
}    

std::cout << "Height: " << tree->getHeight(tree->getRoot()) << std::endl;

delete tree;

return 0;
}

The header file:

#define MAX(a, b) ({ __typeof__ (a) _a = (a); \
                    __typeof__ (b) _b = (b); \
                    _a > _b ? _a : _b; })

template <typename T>
class node
{
    T data;
    node<T> *left;
    node<T> *right;
    node<T> *parent;

public:
    node(T value);
    ~node()  {};
    T getData();
    void updateLeft(node<T>* pointer);
    void updateRight(node<T>* pointer);
    void updateParent(node<T>* pointer);
    node<T>* getLeft();
    node<T>* getRight();
    node<T>* getParent();
};

template <typename T>
node<T> :: node(T value) : parent(NULL), left(NULL), right(NULL)
{
    data = value;
}

template <typename T>
T node<T> :: getData()
{
    return data;
}

template <typename T>
void node<T> :: updateLeft(node<T>* pointer)
{
    left = pointer;
}

template <typename T>
void node<T> :: updateRight(node<T>* pointer)
{
    right = pointer;
}

template <typename T>
void node<T> :: updateParent(node<T>* pointer)
{
    parent = pointer;
}

template <typename T>
node<T>* node<T> :: getLeft()
{
    return left;
}

template <typename T>
node<T>* node<T> :: getRight()
{
    return right;
}

template <typename T>
node<T>* node<T> :: getParent()
{
    return parent;
}

template <typename T>
class BST
{
    node<T> *root;
    void removeNode(node<T> *delNode);
public:
    BST() : root(NULL) {}
    ~BST();
    node<T>* getRoot();
    node<T>* findMin(node<T> *current);
    node<T>* findMax(node<T> *current);
    node<T>* findNode(T key);
    node<T>* getSuccessor(node<T> *current);
    node<T>* getPredecessor(node<T> *current);
    bool addNode(node<T> *newNode);
    void inOrderPrint(node<T> *current);
    void preOrderPrint(node<T> *current);
    void postOrderPrint(node<T> *current);
    void levelOrderPrint(node<T> *current);
    void printLevel(node<T>* current, int level);
    int getHeight(node<T>* current);
};

template <typename T>
void BST<T> :: removeNode(node<T> *delNode)
{
    if(delNode)
    {
        removeNode(delNode->getLeft());
        removeNode(delNode->getRight());
        delete delNode;
    }
}

template <typename T>
BST<T> :: ~BST()
{
    removeNode(root);
}

template <typename T>
node<T>* BST<T> :: getRoot()
{
    return root;
}

template <typename T>
bool BST<T> :: addNode(node<T> *newNode)
{
    if(!root)
    {
        root = newNode;
    }

    else
    {
        node<T> *temp = root;
        node<T> *parent = NULL;

        while(temp)
        {
            parent = std::move(temp);
            if(newNode->getData() > temp->getData())
            {
                temp = temp->getRight();
            }
            else
            {
                temp = temp->getLeft();
            }
        }

        if(newNode->getData() > parent->getData())
        {
            parent->updateRight(newNode);
        }
        else
        {
            parent->updateLeft(newNode);
        }

        newNode->updateParent(parent);

    }
    return true;
}

template <typename T>
node<T>* BST<T> :: findNode(T key)
{
    node<T> *temp = root;
    while(temp)
    {
        T data = temp->getData();
        if(data == key)
        {
            return temp;
        }

        else if(data > key)
        {
            temp = temp->getLeft();
        }

        else
        {
            temp = temp->getRight();
        }
    }

    return NULL;
}

template <typename T>
void BST<T> :: inOrderPrint(node<T> *current)
{
    if(current)
    {
        inOrderPrint(current->getLeft());
        std::cout << current->getData() << " ";
        inOrderPrint(current->getRight());
    }
}

template <typename T>
void BST<T> :: preOrderPrint(node<T> *current)
{
    if(current)
    {
        std::cout << current->getData() << " ";
        inOrderPrint(current->getLeft());
        inOrderPrint(current->getRight());
    }
}

template <typename T>
void BST<T> :: postOrderPrint(node<T> *current)
{
    if(current)
    {
        inOrderPrint(current->getLeft());
        inOrderPrint(current->getRight());
        std::cout << current->getData() << " ";
    }
}

template <typename T>
void BST<T> :: printLevel(node<T>* current, int level)
{
    if(current)
    {
        if(level == 0)
        {
            std::cout << current->getData() << " ";
        }
        else
        {
            printLevel(current->getLeft(), level-1);
            printLevel(current->getRight(), level-1);
        }
    }
}

template <typename T>
void BST<T> :: levelOrderPrint(node<T> *current)
{
    int height = getHeight(current);
    for(int level = 0; level < height; level++)
    {
        printLevel(current, level);
        std::cout << std::endl;
    }
}

template <typename T>
node<T>* BST<T> :: findMin(node<T> *current)
{
    node<T> *temp = current;
    node<T> *prev;
    while(temp)
    {
        prev = temp;
        temp = temp->getLeft();
    }

    return prev;
}

template <typename T>
node<T>* BST<T> :: findMax(node<T> *current)
{
    node<T> *temp = root;
    node<T> *prev;
    while(temp)
    {
        prev = temp;
        temp = temp->getRight();
    }

    return prev;
}

template <typename T>
node<T>* BST<T> :: getSuccessor(node<T> *current)
{
    if(current->getRight())
    {
        return findMin(current->getRight());
    }

    node<T> *parent = current->getParent();
    while(parent && current == parent->getRight())
    {
        current = parent;
        parent = parent->getParent();
    }

    return parent;
}

template <typename T>
node<T>* BST<T> :: getPredecessor(node<T> *current)
{
    if(current->getLeft())
    {
        return findMax(current->getLeft());
    }

    node<T> *parent = current->getParent();
    while(parent && current == parent->getLeft())
    {
        current = parent;
        parent = parent->getParent();
    }

    return parent;
}

template <typename T>
int BST<T> :: getHeight(node<T> *current)
{
    int max = 0;
    if(!current)
    {
        return 0;
    }

    max = MAX(getHeight(current->getLeft()), getHeight(current->getRight()));
    return (1 + max);
}
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  • \$\begingroup\$ We cannot assist with adding smart pointers, but we can review everything else (if it all works). \$\endgroup\$ – Jamal Nov 25 '14 at 23:38
  • \$\begingroup\$ Everything works. Just wanted to know if I'm following best practices. \$\endgroup\$ – user3666471 Nov 25 '14 at 23:39
  • \$\begingroup\$ Can you use / your compiler supports C++11? \$\endgroup\$ – glampert Nov 26 '14 at 1:07
  • 2
    \$\begingroup\$ Yes, I can use C++11. \$\endgroup\$ – user3666471 Nov 26 '14 at 2:58
  • \$\begingroup\$ This will be a fun review. I cannot do it now, but I'll do it in 8ish hours if it has not been done by then. \$\endgroup\$ – nwp Nov 26 '14 at 9:57
3
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No complete review, but some points to mention (in no specific order). When I speak in imperatives, consider them always as suggestions:


General things:

  • Don't use this macro:

        #define MAX(a, b) ({ __typeof__ (a) _a = (a); \
                __typeof__ (b) _b = (b); \
                _a > _b ? _a : _b; })
    

    Rather, use the std::max function template defined in the algorithm header:

        std::max(a,b); 
    
  • Use nullptr instead of NULL.


Smart pointers:

Regarding the design of your node class, here are two suggestions, one using std::unique_ptr the other std::shared_ptr.

Using std::unique_ptr:

First, the unique_ptr alternative. For this, you can use a node class as follows:

    template <typename T>
    class node
    {
        T data;
        node *parent; 
        // or: 
        // node * const parent; 
        std::unique_ptr<node> right;
        std::unique_ptr<node> left;
    }

Note that I didn't use node<T> but rather only node inside the class. This is because inside a class template, the class name refers to its current instantiation.

  • Choice of left and right:

    The child nodes left and right now are unique_ptr's, which states that the node exclusively owns its childs. This is convenient, because, say, when you remove the node from the tree, you usually want all child and grand-child nodes below that erased node to disappear as well. By using std::unique_ptr, you get this for free. Once more, this is the advantage in following the RAII principle: the class manage it's memory ressources itself, so you don't have to care for that manually (as you do via the function removeNode). When you use raw pointers instead, this will become much more difficult and error-prone.

  • Choice of parent:

    On the other hand, parent is chosen as a raw pointer with the intention to act solely as an observer. It relies on the fact that the node object of which it is a member is also correctly managed by a unique_ptr. So it will always point to a valid object. Note that you should never do some memory-related operations using parent.

    As it is also unlikely that child nodes have to change its parent (and since strictly speaking it is forbidden without also changing the managing unique_ptr), you can also declare the raw pointer as const by node * const parent;. Note that the const here comes after the * which means the pointer itself is const (not the object it points to). But changing parent (in a reasonable way) might also be needed in in rare occasions, so it is maybe too restrictive to generally suggest this.

  • Copying, assignment, etc.

    Note that you can't copy the tree with nodes as defined above (at least you won't get it for free, i.e. the copy constructor is not automatically generated by the compiler), which is because std::unique_ptr's copy constructor is delete'd. So there will always exist only one valid tree (which is usually passed by reference). If you want copies, you can either implement a deep copy, by which you get a completely new tree (--if you change the copy, the original tree will not change). Or you can return an observer_node class which resembles yours above (in which you should never do any memory-related operations such as delete):

    template <typename T>
    class observer_node
    {
        T data;
        node<T> *left;
        node<T> *right;
        node<T> *parent;
    };
    

    But I doubt that I really would do this. The copied-node is read-only (with regard to the topology, i.e. you cannot add or remove nodes) and, in order to work, it poses severe restrictions on the original tree. For instance, one cannot reduce the number of nodes, as then the copied-node will contain a dangling pointer.

    For the assignment-- which is usually implemented via the copy-and-swap idiom and thus in terms of the copy constructor -- similar things hold.

    So, if you really want to be able to do pointer-copying, you should consider the next alternative.


Using std::shared_ptr:

Here the class looks like this:

    template <typename T>
    class node : public std::enable_shared_from_this<node<T> >
    {
        T data;
        node *parent;
        // or:
        // std::weak_ptr<node> parent
        std::shared_ptr<node> right;
        std::shared_ptr<node> left;
    }
  • Choice of left and right.

    Now, the child node pointers are shared_ptr. They do not exclusively own the object they point to.

  • Copying, etc:

    The implications are best worked out by considering the copy behaviour (for whose accomplishment you will get compiler-generated copy, assignment, etc. constructors). Now, copying works as following: a new tree of nodes is generated which point to the same memory locations on the heap. In the copied tree, you can safely change the topology of the tree, i.e. add or remove nodes -- the shared_ptr will always take care that the pointed-to objects survive as long as at least one node is pointing to them. You must be careful, though, when you change the member data in the copied tree, as this will change also the value in the original tree.

  • Choice of parent

    parent again can be chosen as an only-observing raw pointer. The rationale is the same way as before: it relies on the fact that the parent node is managed by another shared_ptr and thus will always exist.

    If you wanted to be sure about that, you could also use a weak_ptr here, but I don't think it is necessary.


Oh man, my reviews always get so long ... hope it is helpful and that you even managed to get here.

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  • \$\begingroup\$ One might want to be able to split a tree in two or merge two trees into a larger one. Then the parent pointer would need updating. \$\endgroup\$ – Goswin von Brederlow Nov 26 '14 at 16:55
  • \$\begingroup\$ With std::shared_ptr I would use std::enable_shared_from_this for nodes. That way a node an return a shared pointer to itself to return a subtree without having to go through the parent. This avoids having to special case the root and checking if you need the left or right child. \$\endgroup\$ – Goswin von Brederlow Nov 26 '14 at 16:58
  • \$\begingroup\$ @GoswinvonBrederlow: yes, deriving from std::enable_shared_from_this is a reasonable thing, added this. I had the same feeling about the parent pointer, so I recommended it to be non-const in general. \$\endgroup\$ – davidhigh Nov 26 '14 at 17:03
  • \$\begingroup\$ Small error, std::max is actually defined by <algorithm> not by <cmath> as you have stated. \$\endgroup\$ – glampert Nov 26 '14 at 17:09
  • \$\begingroup\$ @glampert: right, thanks for correction. This is something I quite often stumbled upon... and were always surprised. Corrected. \$\endgroup\$ – davidhigh Nov 26 '14 at 17:14
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I assume you read the other answer already.

  • Your MAX macro is inefficient because it copies its arguments. std::max does not have that problem. Also decltype is a portable version of __typeof__ and in this case you could have used auto instead.

  • int main(){
        BST<int> *tree = new BST<int>;
        ...
        delete tree;
    }
    

    Putting variables on the stack makes managing memory easier than with smart pointers:

    int main(){
        BST<int> tree;
        ...
    }
    

    Exception safety, less code and less awkward syntax are an additional bonus.

  • In main you define int i; but never use it. Your compiler should warn you about that. Define variables as late as possible instead of defining all variables at the beginning of a function.

  • As a user of your class I do not care about nodes. If I have to handle nodes I may as well write my own tree instead of using yours. node is an implementation detail.

    class BST{
        struct Node{
            T data;
            ...
        };
    public:
        ...
    };
    

    Now Node is private and not accessible for users. Instead of bool addNode(node<T> *newNode); we now have void insert(const T &t); and void insert(T &&t);. If the && confuses you read about move semantics. Internally it does basically the same thing as addNode did before. The member functions you have become private and we add new member functions:

    template <class T>
    class BST{
        ...
    public:
        const T &min() const;
        const T &max() const;
        bool contains(const T &key) const;
        void inOrderPrint() const;
        void preOrderPrint() const;
        void postOrderPrint() const;
        void levelOrderPrint() const;
        int getHeight() const;
        void remove(const T &t);
        ...
    };
    

    Those functions also do what they did before, just more convenient to use.

  • BST<int> *tree = new BST<int>;
    tree->addNode(new node<int>(6));
    tree->addNode(new node<int>(5));
    tree->addNode(new node<int>(2));
    tree->addNode(new node<int>(5));
    tree->addNode(new node<int>(7));
    tree->addNode(new node<int>(8));
    

    This sucks. You want it to look something like this:

    BST<int> tree = { 6, 5, 2, 5, 7, 8 };
    

    To do that you add a constructor that takes an initializer_list and inserts all the elements into the tree:

    BST(const std::initializer_list<T> &il){
        for (auto &e : il)
            insert(e);
    }
    BST(std::initializer_list<T> &&il){
        for (auto &e : il)
            insert(std::move(e));
    }
    

    You need to #include <initializer_list> for this to work.

  • I would add copy assignment operators and assignment operators for initializer_list:

    BST &operator =(BST &&other);
    BST &operator =(const BST &other);
    BST &operator =(std::initializer_list<T> &&il);
    BST &operator =(const std::initializer_list<T> &il);
    
  • Consider implementing iterators. An iterator is similar to a pointer, you get the beginning with auto it = bst.begin(); and the next element with ++it;. When it == bst.end(); you are done iterating. The point is that you can separate algorithms such as inOrderPrint and containers such as BST. Instead of implementing n algorithms for m containers with a total of n * m implementations you need only n + m implementations if you use iterators and all the neat algorithms such as std::find_if and std::accumulate will work on your BST. It is quite a bit of work to implement iterators, especially since you will probably need to implement 3 kinds for in-order, pre-order and post-order. Once you have those iterators and add support for streams your tree becomes fun to work with:

    std::ifstream fin("myTree.txt");
    BST<int> bst;
    fin >> bst; //read tree from file
    std::cout << bst; //print tree
    //dumping tree to disk
    std::ofstream fout("preordertree.txt");
    for (auto it = bst.preOrderBegin(); it != bst.preOrderEnd(); ++it)
        fout << *it << '\n';
    

When you get bored adding features to your BST take a close look at std::set and std::vector + std::sort + std::lower_bound.

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  • \$\begingroup\$ Nice complimentary review, +1. Regarding the initializer_list stuff: although it is a good thing in general, I think that tree = { 6, 5, 2, 5, 7, 8 } is not really intuitive: in order to know the distribution of values to the nodes I need to know powers of two. There might be applications where you simply insert somewhere all values you need, but in others this distribution matters, and then the initializer_list is error-prone ... I'm indecisive whether I would add it, with a slight tendency to yes (--lastly we do C++ and users have to know what they do ;-)). Just wanted to mention that. \$\endgroup\$ – davidhigh Nov 27 '14 at 1:08
  • \$\begingroup\$ Next, can you comment what you mean exactly by inOrder, preOrder and postOrder iterators? What exactly does ++ for them? Basically, I can think of several alternatives to traverse the tree, with "layer-by-layer from left-to-right" being probably the easiest to understand. Which one did you mean? \$\endgroup\$ – davidhigh Nov 27 '14 at 1:22
  • \$\begingroup\$ @davidhigh I did not understand the power of two part. When iterating through values the values come in some order, usually in sequential order for vector-like containers or from lowest to highest for set-like containers. The question contained code for inOrderPrint, preOrderPrint, postOrderPrint and levelOrderPrint which one could mimic with different iterator types. Operator ++ would then get the next element based on the order encoded in its type. \$\endgroup\$ – nwp Nov 27 '14 at 2:05
  • \$\begingroup\$ the power-of-two part was meant like this: tree = {1} means there is only the root node with a value of 1. tree = {1,2,3} means there are further left and right childs of the root node. tree = {1,2} means only left is assigned (--first question: is this really desired? often either none or both children are assigned). When I now want to assign,say, 7 to the root->left->left->right node, I must think a while where to insert a 7 in the initializer_list. \$\endgroup\$ – davidhigh Nov 27 '14 at 2:18
  • \$\begingroup\$ this touches also my other question about iterators: in order to use an initializer_list, one needs an intuitive way to traverse the storage scheme (--the same holds for an iterator). So, does tree={1,2,3} means root, root->left, root->right are assigned, or does it mean root, root->left, root->left->left, or something completely different? These questions must be answered in order to use both concepts (iterators and initializer_lists) unambiguously. \$\endgroup\$ – davidhigh Nov 27 '14 at 2:22

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