Balanced Binary Tree - Leetcode 110.py

class Solution:
    def isBalanced(self, root: Optional[TreeNode]) -> bool:
        balanced = [True]

        def height(root):
            if not root:
                return 0

            left_height = height(root.left)
            if balanced[0] is False:
                return 0


            right_height = height(root.right)
            if abs(left_height - right_height) > 1:
                balanced[0] = False
                return 0
            return 1 + max(left_height, right_height)

        height(root)
        return balanced[0]

# Time Complexity: O(n)
# Space Complexity: O(h) { here "h" is the height of the tree }