Check if a Binary Tree is Balanced

Algorithm Steps

Given a binary tree, if the tree is empty, it is considered balanced.
Recursively compute the height of the left and right subtrees.
If the absolute difference between the left and right subtree heights is more than 1, the tree is not balanced.
Recursively check that both the left subtree and the right subtree are balanced.
If all checks pass, the binary tree is balanced.

Check if a Binary Tree is Balanced - Code Examples Code

Python

Java

JavaScript

C++

Kotlin

Swift

Php

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


def isBalanced(root):
    def height(node):
        if not node:
            return 0
        return max(height(node.left), height(node.right)) + 1
    if not root:
        return True
    leftHeight = height(root.left)
    rightHeight = height(root.right)
    if abs(leftHeight - rightHeight) > 1:
        return False
    return isBalanced(root.left) and isBalanced(root.right)

# Example usage:
if __name__ == '__main__':
    # Construct a binary tree:
    #         1
    #        / \
    #       2   3
    #      /
    #     4
    root = TreeNode(1, TreeNode(2, TreeNode(4)), TreeNode(3))
    print(isBalanced(root))