Binary Tree Tilt Python Solution with Tests

lcpy gen -n 563   # by problem number
lcpy gen -s binary_tree_tilt   # by problem name

Input: root = [1,2,3]
Output: 1
Explanation:
Tilt of node 2 : |0-0| = 0 (no children)
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 1 : |2-3| = 1 (left subtree is just left child, so sum is 2; right subtree is just right child, so sum is 3)
Sum of every tilt : 0 + 0 + 1 = 1

Input: root = [4,2,9,3,5,null,7]
Output: 15
Explanation:
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 5 : |0-0| = 0 (no children)
Tilt of node 7 : |0-0| = 0 (no children)
Tilt of node 2 : |3-5| = 2 (left subtree is just left child, so sum is 3; right subtree is just right child, so sum is 5)
Tilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7)
Tilt of node 4 : |(3+5+2)-(9+7)| = |10-16| = 6 (left subtree values are 3, 5, and 2, which sums to 10; right subtree values are 9 and 7, which sums to 16)
Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15

Input: root = [21,7,14,1,1,2,2,3,3]
Output: 9

from leetcode_py import TreeNode


class Solution:
    # Time: O(n)
    # Space: O(h) for the recursion stack
    def find_tilt(self, root: TreeNode[int] | None) -> int:
        total = 0

        def dfs(node: TreeNode[int] | None) -> int:
            nonlocal total
            if node is None:
                return 0
            left_sum = dfs(node.left)
            right_sum = dfs(node.right)
            total += abs(left_sum - right_sum)
            return left_sum + right_sum + node.val

        dfs(root)
        return total