数组遍历中的BST

Question

I have the following implementation of a binary tree in an array;

   32
  /  \
 2    -5
     /  \
   -331   399

The data is grouped 3 indexes at a time. index%3==0 is the value of the node, index%3==1 is the index of the value of the left node and index%3==2 is the index of the value of the right node. If the left or right index reference is 0, there is no node that direction.

I'm trying to find the depth (height) of this tree. I've written it recursively

height(node): 
   if node == null:
        return 0
   else:
        return max(height(node.L), height(node.R)) + 1

I want to find a non-recursive solution, however.

Here is some pseudocode i have, assuming the tree is not empty

int i = 0; int left = 0; int right = 0;
while (i != n ){
if ( a[i+1] != 0 ){
  left++;
}
else if ( a[i+2] != 0 ){
  right++;
}
 i = i + 3;
 }

return max ( left, right ) + 1;

I don't think this is right and I'd like some help figuring out how to do this correctly.

Answer 1

You haven't said what your problem is with recursion for us to understand what behavior you want to improve.

There are many solutions to this, but almost all of them have the same or worse performance than your recursive solution. Really, the best solutions are going to be things you'd have to do when you're creating the tree. For example, you could store the height of each node in a fourth array index per node. Then it's a trivial scan of every fourth index to find the max height. It would also make it easier if nodes had parent references stored with them so that didn't have to be computed during the height check.

One solution is to simulate recursion with a stack, but that's really no different than recursion.

Another solution is to go through each node and determine its height based on it's parent, but not in a specific traversal's order. However, because of how you have this configured, without a secondary datastructure to store the hierarchy, it's going to be less efficient O(n^2). The problem is you can't get from the child to its parent without a full array scan. Then you can do it in linear time (but recursion is also linear time, so I'm not sure we're doing better. It's also not going to be much better from a memory perspective).

Can you define what type of efficiency you want to improve?

Here's the pseudocode for each, but I'm depending on a few datastructures that aren't easily present:

"recursion without recursion" solution:

int get_height(int * tree, int length) {

    Stack stack;

    int max_height = 0;

    if (length == 0) {
        return 0;
    }

    // push an "array" of the node index to process and the height of its parent.  
    //   make this a struct and use that for real c code
    stack.push(0,0);

    while(!stack.empty()) {
        int node_index, parent_height = stack.pop();

        int height = parent_height + 1;
        if (height > max_height) {
            max_height=height;
        }
        if (tree[node_index+1] != 0 )
            stack.push(tree[node_index+1], height);
        if (tree[node_index+2] != 0 )
            stack.push(tree[node_index+2], height);

    }

    return max_height;
}

Now working on really slow solution that uses no additional memory, but it's REALLY bad. It's like writing fibonacci recursively bad. The original algorithm went through each node and performed O(n) checks worst case for a runtime of O(n^2) (actually not quite as bad as I had originally thought)

edit: much later I'm adding an optimization that skips all nodes with children. This is REALLY important, as it cuts out a lot of calls. Best case is if the tree is actually a linked list, in which case it runs in O(n) time. Worst case is a fully balanced tree - with logn leaf nodes each doing logn checks back to the root for O((log(n)^2). Which isn't nearly so bad. Lines below to be marked as such

"really slow but no extra memory" solution (but now updated to not be nearly so slow):

int get_height(int * tree, int length) {
    int max_height = 0;
    for (int i = 0; i < length; i+=3) {

        // Optimization I added later
        // if the node has children, it can't be the tallest node, so don't
        //   bother checking from here, as the child will be checked
        if (tree[i+1] != 0 || tree[i+2] != 0)
            continue;

        int height = 0;
        int index_pointing_at_me;

        // while we haven't gotten back to the head of the tree, keep working up
        while (index_pointing_at_me != 0) {
            height += 1; 
            for (int j = 0; j < length; j+=3) {
                if (tree[j+1] == tree[i] ||
                    tree[j+2] == tree[i]) {
                    index_pointing_at_me = j;
                    break;
                }
            }

        }
        if (height > max_height) {
            max_height = height;
        }

    }

    return max_height;
}

Improved on previous solution, but uses O(n) memory - this assumes parents are always before children in array (which I suppose isn't technically required)

int get_height(int * tree, int length) {

    if (length == 0) 
        return 0;

    // two more nodes per node - one for which node is its parent, the other for its height
    int * reverse_mapping = malloc((sizeof(int) * length / 3) * 2) 
    reverse_mapping[1] = 1; // set height to 1 for first node



    // make a mapping from each node to the node that points TO it.
    // for example, for the first node
    //    a[0] = 32
    //    a[1] = 3
    //    a[2] = 6
    //  store that the node at 3 and 6 are both pointed to by node 0 (divide by 3 just saves space since only one value is needed) and that each child node is one taller than its parent
    int max_height = 0;
    for (int i = 0; i < length; i+=3) {

        int current_height = reverse_mapping[(i/3)*2+1];
        if (current_height > max_height)
            max_height = current_height;

        reverse_mapping[(tree[i+1]/3)*2] = i;
        reverse_mapping[(tree[i+1]/3)*2 + 1] = current_height + 1;

        reverse_mapping[(tree[i+2]/3)*2] = i;
        reverse_mapping[(tree[i+2]/3)*2 + 1] = current_height + 1;

    }
    return max_height
}