Check for Balanced Tree

Problem Statement - link #

Given a binary tree, find if it is height balanced or not. A tree is height balanced if difference between heights of left and right subtrees is not more than one for all nodes of tree.

A height balanced tree
        1
     /     \
   10      39
  /
5

An unbalanced tree
        1
     /    
   10   
  /
5    

Your Task: You don’t need to take input. Just complete the function isBalanced() that takes root node as parameter and returns true, if the tree is balanced else returns false.

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(h) where h is height of tree

Examples #

Example 1:

Input:
       10
     /   \
    20   30 
  /   \
 40   60
Output: 1
Explanation: The max difference in height
of left subtree and right subtree is 1.
Hence balanced. 

Example 2:

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

Constraints #

• 1 <= Number of nodes <= 10^5
• 0 <= Data of a node <= 10^6

Solutions #

class Solution{
    public:
    //Function to check whether a binary tree is balanced or not.
    int height(Node* root){
        if(!root) return 0;
        int lh = height(root->left);
        if(lh==-1) return -1;
        int rh = height(root->right);
        if(rh==-1) return -1;
        if(abs(lh-rh)>1) return -1;
        return max(lh,rh)+1;
    }
    bool isBalanced(Node *root)
    {
        //  Your Code here
        return height(root)!=-1;
    }
};