**Height Balanced Tree - Amazon Top Interview Questions**

### Problem Statement :

Given the root of a binary tree, return whether its height is balanced. That is, for every node in the tree, the absolute difference of the height of its left subtree and the height of its right subtree is 0 or 1. Constraints n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [1, null, [4, null, [12, null, null]]] Output False Explanation This is false since the root's right subtree has height of 2, and left has height of 0. Example 2 Input root = [1, [0, null, null], [4, null, [12, null, null]]] Output True

### Solution :

` ````
Solution in C++ :
int height(Tree* root) {
if (root == NULL) return 0;
return max(height(root->left), height(root->right)) + 1;
}
bool solve(Tree* root) {
if (root == NULL) return true;
return solve(root->left) && solve(root->right) &&
abs(height(root->left) - height(root->right)) <= 1;
}
```

` ````
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
public boolean ans = true;
public int traverse(Tree root) {
if (root == null)
return 0;
int left = traverse(root.left);
int right = traverse(root.right);
ans = ans && (Math.abs(left - right) < 2);
return Math.max(left, right) + 1;
}
public boolean solve(Tree root) {
traverse(root);
return ans;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, root):
isBalanced = True
def dfs(depth, root):
nonlocal isBalanced
if not root:
return depth
left = dfs(depth + 1, root.left)
right = dfs(depth + 1, root.right)
if abs(left - right) > 1:
isBalanced = False
return max(left, right)
dfs(0, root)
return isBalanced
```

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