The best way to calculate the height in a binary search tree? (balancing an AVL-tree)

Question

I\'m looking for the best way to calculate a nodes balance in an AVL-tree. I thought I had it working, but after some heavy inserting/updating I can see that it\'s not worki

Answer 1

Part 1 - height

As starblue says, height is just recursive. In pseudo-code:

height(node) = max(height(node.L), height(node.R)) + 1

Now height could be defined in two ways. It could be the number of nodes in the path from the root to that node, or it could be the number of links. According to the page you referenced, the most common definition is for the number of links. In which case the complete pseudo code would be:

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

If you wanted the number of nodes the code would be:

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

Either way, the rebalancing algorithm I think should work the same.

However, your tree will be much more efficient (O(ln(n))) if you store and update height information in the tree, rather than calculating it each time. (O(n))

Part 2 - balancing

When it says "If the balance factor of R is 1", it is talking about the balance factor of the right branch, when the balance factor at the top is 2. It is telling you how to choose whether to do a single rotation or a double rotation. In (python like) Pseudo-code:

if balance factor(top) = 2: // right is imbalanced
     if balance factor(R) = 1: // 
          do a left rotation
     else if balance factor(R) = -1:
          do a double rotation
else: // must be -2, left is imbalanced
     if balance factor(L) = 1: // 
          do a left rotation
     else if balance factor(L) = -1:
          do a double rotation

I hope this makes sense

Answer 2

This BFS-like solution is pretty straightforward. Simply jumps levels one-by-one.

def getHeight(self,root, method='links'):
    c_node = root
    cur_lvl_nodes = [root]
    nxt_lvl_nodes = []
    height = {'links': -1, 'nodes': 0}[method]

    while(cur_lvl_nodes or nxt_lvl_nodes):
        for c_node in cur_lvl_nodes:
            for n_node in filter(lambda x: x is not None, [c_node.left, c_node.right]):
                nxt_lvl_nodes.append(n_node)

        cur_lvl_nodes = nxt_lvl_nodes
        nxt_lvl_nodes = []
        height += 1

    return height

Answer 3

Give BinaryTree<T, Comparator>::Node a subtreeHeight data member, initialized to 0 in its constructor, and update automatically every time with:

template <typename T, typename Comparator>
inline void BinaryTree<T, Comparator>::Node::setLeft (std::shared_ptr<Node>& node) {
    const std::size_t formerLeftSubtreeSize = left ? left->subtreeSize : 0;
    left = node;
    if (node) {
        node->parent = this->shared_from_this();
        subtreeSize++;
        node->depthFromRoot = depthFromRoot + 1;
        const std::size_t h = node->subtreeHeight;
        if (right)
            subtreeHeight = std::max (right->subtreeHeight, h) + 1;
        else
            subtreeHeight = h + 1;
    }
    else {
        subtreeSize -= formerLeftSubtreeSize;
        subtreeHeight = right ? right->subtreeHeight + 1 : 0;
    }
}

template <typename T, typename Comparator>
inline void BinaryTree<T, Comparator>::Node::setRight (std::shared_ptr<Node>& node) {
    const std::size_t formerRightSubtreeSize = right ? right->subtreeSize : 0;
    right = node;
    if (node) {
        node->parent = this->shared_from_this();
        subtreeSize++;
        node->depthFromRoot = depthFromRoot + 1;
        const std::size_t h = node->subtreeHeight;
        if (left)
            subtreeHeight = std::max (left->subtreeHeight, h) + 1;
        else
            subtreeHeight = h + 1;
    }
    else {
        subtreeSize -= formerRightSubtreeSize;
        subtreeHeight = left ? left->subtreeHeight + 1 : 0;
    }
}

Note that data members subtreeSize and depthFromRoot are also updated. These functions are called when inserting a node (all tested), e.g.

template <typename T, typename Comparator>
inline std::shared_ptr<typename BinaryTree<T, Comparator>::Node>
BinaryTree<T, Comparator>::Node::insert (BinaryTree& tree, const T& t, std::shared_ptr<Node>& node) {
    if (!node) {
        std::shared_ptr<Node> newNode = std::make_shared<Node>(tree, t);
        node = newNode;
        return newNode;
    }
    if (getComparator()(t, node->value)) {
        std::shared_ptr<Node> newLeft = insert(tree, t, node->left);
        node->setLeft(newLeft);
    }
    else {
        std::shared_ptr<Node> newRight = insert(tree, t, node->right);
        node->setRight(newRight);
    }
    return node;
}

If removing a node, use a different version of removeLeft and removeRight by replacing subtreeSize++; with subtreeSize--;. Algorithms for rotateLeft and rotateRight can be adapted without much problem either. The following was tested and passed:

template <typename T, typename Comparator>
void BinaryTree<T, Comparator>::rotateLeft (std::shared_ptr<Node>& node) {  // The root of the rotation is 'node', and its right child is the pivot of the rotation.  The pivot will rotate counter-clockwise and become the new parent of 'node'.
    std::shared_ptr<Node> pivot = node->right;
    pivot->subtreeSize = node->subtreeSize;
    pivot->depthFromRoot--;
    node->subtreeSize--;  // Since 'pivot' will no longer be in the subtree rooted at 'node'.
    const std::size_t a = pivot->left ? pivot->left->subtreeHeight + 1 : 0;  // Need to establish node->heightOfSubtree before pivot->heightOfSubtree is established, since pivot->heightOfSubtree depends on it.
    node->subtreeHeight = node->left ? std::max(a, node->left->subtreeHeight + 1) : std::max<std::size_t>(a,1);
    if (pivot->right) {
        node->subtreeSize -= pivot->right->subtreeSize;  // The subtree rooted at 'node' loses the subtree rooted at pivot->right.
        pivot->subtreeHeight = std::max (pivot->right->subtreeHeight, node->subtreeHeight) + 1;
    }
    else
        pivot->subtreeHeight = node->subtreeHeight + 1;
    node->depthFromRoot++;
    decreaseDepthFromRoot(pivot->right);  // Recursive call for the entire subtree rooted at pivot->right.
    increaseDepthFromRoot(node->left);  // Recursive call for the entire subtree rooted at node->left.
    pivot->parent = node->parent;
    if (pivot->parent) {  // pivot's new parent will be its former grandparent, which is not nullptr, so the grandparent must be updated with a new left or right child (depending on whether 'node' was its left or right child).
        if (pivot->parent->left == node)
            pivot->parent->left = pivot;
        else
            pivot->parent->right = pivot;
    }
    node->setRightSimple(pivot->left);  // Since pivot->left->value is less than pivot->value but greater than node->value.  We use the NoSizeAdjustment version because the 'subtreeSize' values of 'node' and 'pivot' are correct already.
    pivot->setLeftSimple(node);
    if (node == root) {
        root = pivot;
        root->parent = nullptr; 
    }
}

where

inline void decreaseDepthFromRoot (std::shared_ptr<Node>& node) {adjustDepthFromRoot(node, -1);}
inline void increaseDepthFromRoot (std::shared_ptr<Node>& node) {adjustDepthFromRoot(node, 1);}

template <typename T, typename Comparator>
inline void BinaryTree<T, Comparator>::adjustDepthFromRoot (std::shared_ptr<Node>& node, int adjustment) {
    if (!node)
        return;
    node->depthFromRoot += adjustment;
    adjustDepthFromRoot (node->left, adjustment);
    adjustDepthFromRoot (node->right, adjustment);
}

Here is the entire code: http://ideone.com/d6arrv

Answer 4

Here's where it gets confusing, the text states "If the balance factor of R is 1, it means the insertion occurred on the (external) right side of that node and a left rotation is needed". But from m understanding the text said (as I quoted) that if the balance factor was within [-1, 1] then there was no need for balancing?

R is the right-hand child of the current node N.

If balance(N) = +2, then you need a rotation of some sort. But which rotation to use? Well, it depends on balance(R): if balance(R) = +1 then you need a left-rotation on N; but if balance(R) = -1 then you will need a double-rotation of some sort.

Answer 5

Here's an alternate way of finding height. Add an additional attribute to your node called height:

class Node
{
data value; //data is a custom data type
node right;
node left;
int height;
}

Now, we'll do a simple breadth-first traversal of the tree, and keep updating the height value for each node:

int height (Node root)
{
Queue<Node> q = Queue<Node>();
Node lastnode;
//reset height
root.height = 0;

q.Enqueue(root);
while(q.Count > 0)
{
   lastnode = q.Dequeue();
   if (lastnode.left != null){
      lastnode.left.height = lastnode.height + 1; 
      q.Enqueue(lastnode.left);
   }

   if (lastnode.right != null){
      lastnode.right.height = lastnode.height + 1;
      q.Enqueue(lastnode.right);
   }
}
return lastnode.height; //this will return a 0-based height, so just a root has a height of 0
}

Cheers,

Answer 6

Here's where it gets confusing, the text states "If the balance factor of R is 1, it means the insertion occurred on the (external) right side of that node and a left rotation is needed". But from m understanding the text said (as I quoted) that if the balance factor was within [-1, 1] then there was no need for balancing?

Okay, epiphany time.

Consider what a rotation does. Let's think about a left rotation.

 P = parent
 O = ourself (the element we're rotating)
 RC = right child
 LC = left child (of the right child, not of ourself)

 P              10
  \               \
   O               15
    \                \
     RC               20
    /                /
   LC               18

          ↓
 P              10
   \               \
     RC              20
   /               /
  O              15
   \               \
   LC              18

 basically, what happens is;

 1. our right child moves into our position
 2. we become the left child of our right child
 3. our right child's left child becomes our right

Now, the big thing you have to notice here - this left rotation HAS NOT CHANGED THE DEPTH OF THE TREE. We're no more balanced for having done it.

But - and here's the magic in AVL - if we rotated the right child to the right FIRST, what we'd have is this...

 P
  \
   O
    \
     LC
      \
       RC

And NOW if we rotate O left, what we get is this...

 P
   \
     LC
    /  \
   O   RC

Magic! we've managed to get rid of a level of the tree - we've made the tree balanced.

Balancing the tree means getting rid of excess depth, and packing the upper levels more completely - which is exactly what we've just done.

That whole stuff about single/double rotations is simply that you have to have your subtree looking like this;

 P
  \
   O
    \
     LC
      \
       RC

before you rotate - and you may have to do a right rotate to get into that state. But if you're already in that state, you only need to do the left rotate.