In case of binary trees, if the trees are skewed, they become computationally inefficient to perform operations on.
This is the motivation behind making sure that trees are not skewed. Hence the need for balanced binary trees.
Balanced Binary trees are computationally efficient to perform operations on.
A balanced binary tree will follow the following conditions:
Balanced binary trees are also known as height-balanced binary trees. Height balanced binary trees can be denoted by HB(k), where k is the difference between heights of left and right subtrees. ‘k’ is known as the balance factor.
If for a tree, the balance factor (k) is equal to zero, then that tree is known as a fully balanced binary tree. It can be denoted as HB(0).
If a binary search tree has a balance factor of one then it is an AVL ( Adelso-Velskii and Landis) tree. This means that in an AVL tree the difference between left subtree and right subtree height is at most one.
AVL tree is a self-balancing binary search tree. In an AVL tree if the difference between left and right subtrees is greater than 1 then it performs one of the following 4 rotations to rebalance itself :
To check if a Binary tree is balanced we need to check three conditions :
We will need a function that can calculate the height of the tree. One way to do this is to write a separate function for calculating the height and call it every time height is needed. This is going to be computationally inefficient.
The better way to implement this will be returning height in the same function.
For each node, we will return -1 if it is not balanced and the height of that node/subtree if it is balanced.
The algorithm is as follows :
The pseudo code will look like as follows :
private int balanceHeight (TreeNode currentNode)
{
if (currentNode == null)
{
return 0;
}
// checking left subtree
int leftSubtreeHeight = balanceHeight (currentNode.left);
if (leftSubtreeHeight == -1) return -1;
// if left subtree is not balanced then the entire tree is also not balanced
//checking right subtree
int rightSubtreeHeight = balanceHeight (currentNode.right);
if (rightSubtreeHeight == -1) return -1;
// if right subtree is not balanced then the entire tree is also not balanced
//checking the difference of left and right subtree for current node
if (Math.abs(leftSubtreeHeight - rightSubtreeHeight) > 1)
{
return -1;
}
//if it is balanced then return the height
return (Math.max(leftSubtreeHeight, rightSubtreeHeight) + 1);
}
You can call this function on the root of the tree and if it returns anything other than -1 then it is a balanced binary tree. If it returns -1 then it is not a balanced binary tree.
In this tutorial we covered Balanced binary trees and how can we check if a tree is balanced binary tree. Hope you understood and enjoyed learning with us.
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