Exception free tree destruction in C++

Question

I have recently managed to get a stack overflow when destroying a tree by deleting its root 'Node', while the Node destructor is similar to this:

Node::~Node(){
    for(int i=0;i<m_childCount;i++)
        delete m_child[i];
}

A solution that come up into my mind was to use own stack. So deleting the tree this way:

std::stack< Node* > toDelete;

if(m_root)
    toDelete.push(m_root);

while(toDelete.size()){
    Node *node = toDelete.top();
    toDelete.pop();

    for(int i=0;i<node->GetChildCount();i++)
        toDelete.push(node->Child(i));

    delete node;
}

But in there the std::stack::push() may throw an exception. Is it possible to write an exception free tree destruction? How?

EDIT:

If anybody is interested here is an exception free non-recursive code inspired by the algorithm pointed out by jpalecek:

Node *current = m_root;

while(current){
  if(current->IsLeaf()){
    delete current;
    return;
  }

  Node *leftMostBranch = current;// used to attach right subtrees

  // delete all right childs
  for(size_t i=1; i<current->GetChildCount(); i++){
    while(!leftMostBranch->Child(0)->IsLeaf())
      leftMostBranch = leftMostBranch->Child(0);

    delete leftMostBranch->Child(0);
    leftMostBranch->Child(0) = current->Child(i);
  }

  // delete this node and advance to the left child
  Node *tmp = current;
  current = current->Child(0);
  delete tmp;
}

note: Node::IsLeaf() is equivalent to Node::GetChildCount()!=0.

Answer 1

This is what all garbage collectors struggle with. However, the best thing you can do (IMHO) is to pray for enough memory for the stack, and your prayers will be heard 99.99999% of the time. Should it not happen, just abort().

BTW if you are interested, there is a solution to traverse long (and deep) trees without allocating much memory.

Answer 2

Why is the original code throwing an exception? I'm guessing you are doing something like using the same node object in multiple places in the tree. Stack overflows are rarely caused by normal expected situations. Stack overflows are not a problem, they are the symptom of a problem.

Rewriting the code differently won't fix that; you should just investigate & fix the error.

Answer 3

Is it possible to write an exception free tree destruction? How?

Perhaps this (untested code):

void destroy(Node* parent)
{
  while (parent)
  {
    //search down to find a leaf node, which has no children
    Node* leaf = parent;

    while (leaf->children.count != 0)
      leaf = leaf->chilren[0];

    //remember the leaf's parent
    parent = leaf->parent;

    //delete the leaf
    if (parent)
    {
      parent->children.remove(leaf);
    }
    delete leaf; 

  } //while (parent)
}

Answer 4

I just had this as an interview question.

And I must admit this is one of the hardest things I had to solve on the fly.
Personally I don't think it's a good question as you may know the trick (if you have read Knuth) in which case it becomes trivial to solve but you can still fool the interviewer into making him/her think you have solved it on the fly.

This can be done assuming that the node stores child pointers in a static structure. If the node stores child pointers in a dynamic structure then it will not work, as you need to re-shape the tree on the fly (it may work but there is no guarantee).

Surprisingly the solution is O(n)
(Technically every node is visited exactly twice with no re-scanning of the tree).
This solution uses a loop (so no memory usage for stack) and does not dynamically allocate memeroy to hold nodes that need to be deleted. So it is surprisingly effecient.

class Node
{
    // Value we do not care about.
    int   childCount;
    Node* children[MAX_CHILDREN];
 };

freeTree(Node* root)
{
    if (root == NULL)
    {    return;
    }
    Node* bottomLeft  = findBottomLeft(root);

    while(root != NULL)
    {
        // We want to use a loop not recursion.
        // Thus we need to make the tree into a list.
        // So as we hit a node move all children to the bottom left.
        for(int loop = 1;loop < root->childCount; ++loop)
        {
            bottomLeft->children[0] = root->children[loop];
            bottomLeft->childCount  = std::max(1, bottomLeft->childCount);
            bottomLeft = findBottomLeft(bottomLeft);
        }

        // Now we have a root with a single child
        // Now we can delete the node and move to the next node.
        Node* bad = root;
        root      = root->children[0];
        delete bad;     // Note the delete should no longer destroy the children.
    }
}

Node* findBottomLeft(Node* node)
{
    while((node->childCount > 0) && node->children[0] != NULL))
    {    node = node->children[0];
    }
    return node;
}

The above method will work as long as their is always a children[0] node (even if it is NULL). As long as you do not have to dynamically allocate space to hold children[0]. Alternatively just add one more pointer to the node object to hold the delete list and use this to turn the tree into a list.