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 :



title-img 




                        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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