Topic: Tree

1) Check for a balanced binary tree

Easy

Input:
      1
    /
   2
    \
     3 
Output: 0
Explanation: The max difference in height
of left subtree and right subtree is 2,
which is greater than 1. Hence unbalanced

class Tree
{

    int isleftBalanced(Node root)
    {
        if(root==null){return 0;}
        else{
            return Math.max(isleftBalanced(root.left),isleftBalanced(root.right))+1;
        }
    }
    int isrightBalanced(Node root)
    {
        if(root==null){return 0;}
        else{

            return Math.max(isrightBalanced(root.left),isrightBalanced(root.right))+1;
        }
    }


    boolean isBalanced(Node root)
    {
    // TOP - DOWN APPROACH

        if(root==null){
            return true;
        }
        else{
            int leftCount = isleftBalanced(root.left);
            int rightCount = isrightBalanced(root.right);

            return (Math.abs(leftCount-rightCount)<=1 && isBalanced(root.left) && isBalanced(root.right));
        }
    }
}

Thank you for reading :)