How to determine if binary tree is balanced?

Question

It\'s been a while from those school years. Got a job as IT specialist at a hospital. Trying to move to do some actual programming now. I\'m working on binary trees now, a

Answer 1

Bonus exercise response. The simple solution. Obviously in a real implementation one might wrap this or something to avoid requiring the user to include height in their response.

IsHeightBalanced(tree, out height)
    if (tree is empty)
        height = 0
        return true
    balance = IsHeightBalanced(tree.left, heightleft) and IsHeightBalanced(tree.right, heightright)
    height = max(heightleft, heightright)+1
    return balance and abs(heightleft - heightright) <= 1     

Answer 2

Well, you need a way to determine the heights of left and right, and if left and right are balanced.

And I'd just return height(node->left) == height(node->right);

As to writing a height function, read: Understanding recursion

Answer 3

class Node {
    int data;
    Node left;
    Node right;

    // assign variable with constructor
    public Node(int data) {
        this.data = data;
    }
}

public class BinaryTree {

    Node root;

    // get max depth
    public static int maxDepth(Node node) {
        if (node == null)
            return 0;

        return 1 + Math.max(maxDepth(node.left), maxDepth(node.right));
    }

    // get min depth
    public static int minDepth(Node node) {
        if (node == null)
            return 0;

        return 1 + Math.min(minDepth(node.left), minDepth(node.right));
    }

    // return max-min<=1 to check if tree balanced
    public boolean isBalanced(Node node) {

        if (Math.abs(maxDepth(node) - minDepth(node)) <= 1)
            return true;

        return false;
    }

    public static void main(String... strings) {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);


        if (tree.isBalanced(tree.root))
            System.out.println("Tree is balanced");
        else
            System.out.println("Tree is not balanced");
    }
}

Answer 4

The definition of a height-balanced binary tree is:

Binary tree in which the height of the two subtrees of every node never differ by more than 1.

So, An empty binary tree is always height-balanced.
A non-empty binary tree is height-balanced if:

1. Its left subtree is height-balanced.
2. Its right subtree is height-balanced.
3. The difference between heights of left & right subtree is not greater than 1.

Consider the tree:

    A
     \ 
      B
     / \
    C   D

As seen the left subtree of A is height-balanced (as it is empty) and so is its right subtree. But still the tree is not height-balanced as condition 3 is not met as height of left-subtree is 0 and height of right sub-tree is 2.

Also the following tree is not height balanced even though the height of left and right sub-tree are equal. Your existing code will return true for it.

       A
     /  \ 
    B    C
   /      \
  D        G
 /          \
E            H

So the word every in the def is very important.

This will work:

int height(treeNodePtr root) {
        return (!root) ? 0: 1 + MAX(height(root->left),height(root->right));
}

bool isHeightBalanced(treeNodePtr root) {
        return (root == NULL) ||
                (isHeightBalanced(root->left) &&
                isHeightBalanced(root->right) &&
                abs(height(root->left) - height(root->right)) <=1);
}

Ideone Link

Answer 5

Note 1: The height of any sub-tree is computed only once.

Note 2: If the left sub-tree is unbalanced then the computation of the right sub-tree, potentially containing million elements, is skipped.

// return height of tree rooted at "tn" if, and only if, it is a balanced subtree
// else return -1
int maxHeight( TreeNode const * tn ) {
    if( tn ) {
        int const lh = maxHeight( tn->left );
        if( lh == -1 ) return -1;
        int const rh = maxHeight( tn->right );
        if( rh == -1 ) return -1;
        if( abs( lh - rh ) > 1 ) return -1;
        return 1 + max( lh, rh );
    }
    return 0;
}

bool isBalanced( TreeNode const * root ) {
    // Unless the maxHeight is -1, the subtree under "root" is balanced
    return maxHeight( root ) != -1;
}

Answer 6

If this is for your job, I suggest:

1. do not reinvent the wheel and
2. use/buy COTS instead of fiddling with bits.
3. Save your time/energy for solving business problems.