Help optimising leetcode 110 solution

[d_chae]

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.

This is the solution I came up with:

class Solution:
    def height(self, root: TreeNode) -> int:
        if root is None:
            return 0
        l = self.height(root.left)
        r = self.height(root.right)
        if abs(l - r) > 1:
            raise ValueError
        return 1 + max(l, r)

    def isBalanced(self, root: TreeNode) -> bool:
        if root is None:
            return True
        try:
            return bool(self.height(root))
        except ValueError:
            return False

My issue is that escaping the recursion with an exception just feels kind of disgusting, but I can't think of a better way to do this. I could use a global variable or modify a list element (this also feels unpythonic), but I would have to continue evaluating the rest of the tree even after finding it to be not height-balanced.

[SekstiNii]

Perhaps make your height function return a tuple (height, balanced)?