# Balanced Binary Tree – Leetcode Solution

In this post, we are going to solve the 110. Balanced Binary Tree problem of Leetcode. This problem 110. Balanced Binary Tree is a Leetcode easy level problem. Let’s see the code, 110. Balanced Binary Tree – Leetcode Solution.

## Problem

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

• a binary tree in which the left and right subtrees of every node differ in height by no more than 1.

### Example 1 :

``````Input: root = [3,9,20,null,null,15,7]
Output: true``````

### Example 2 :

``````Input: root = [1,2,2,3,3,null,null,4,4]
Output: false``````

### Example 3 :

``````Input: root = []
Output: true``````

### Constraints

• The number of nodes in the tree is in the range `[0, 5000].`
• `-104 <= Node.val <= 104`

Now, let’s see the code of 110. Balanced Binary Tree – Leetcode Solution.

# Balanced Binary Tree – Leetcode Solution

### 110. Balanced Binary Tree – Solution in Java

```/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public int height(TreeNode root){
if(root==null) return 0;

int lht = height(root.left);
int rht = height(root.right);

return Math.max(lht,rht)+1;

}
public boolean isBalanced(TreeNode root) {
if(root == null) return true;
if(root.left == null && root.right == null) return true;

int lht = height(root.left);
int rht = height(root.right);

if(Math.abs(lht-rht)>1){
return false;
}

boolean checkLeft = isBalanced(root.left);
boolean checkRight = isBalanced(root.right);

if(checkLeft == false || checkRight == false) return false;

return true;
}
}```

### 110. Balanced Binary Tree – Solution in C++

```class Solution {
public:
bool ans;
int checkBalance(TreeNode* root){
if(!root)
return 0;
return 0;
int leftSubTree = checkBalance(root->left);
int rightSubTree = checkBalance(root->right);
if(abs(leftSubTree-rightSubTree) > 1){
ans = false;
}
return 1 + max(leftSubTree, rightSubTree);
}
bool isBalanced(TreeNode* root){
ans = true;
int temp = checkBalance(root);
return ans;
}
};```

### 110. Balanced Binary Tree – Solution in Python

```class Solution(object):
def isBalanced(self, root):

def check(root):
if root is None:
return 0
left  = check(root.left)
right = check(root.right)
if left == -1 or right == -1 or abs(left - right) > 1:
return -1
return 1 + max(left, right)

return check(root) != -1```

Note: This problem 110. Balanced Binary Tree is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.