Count Univalue Subtrees

没有太好的办法，只能divide conquer然后汇总，关于空节点和单个结点树的处理非常麻烦

class Solution(object):
    count = 0
    def countUnivalSubtrees(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        self.helper(root)
        return self.count

    def helper(self, root):
        if not root.left and not root.right:
            self.count += 1
            return (True, root.val)

        left, right = True, True
        leftVal, rightVal = None, None
        if root.left:
            left, leftVal = self.helper(root.left)
        if root.right:
            right, rightVal = self.helper(root.right)

        if left and right:
            if leftVal is None and root.val == rightVal:
                self.count += 1
                return (True, root.val)
            if rightVal is None and root.val == leftVal:
                self.count += 1
                return (True, root.val)
            if root.val == leftVal == rightVal:
                self.count += 1
                return (True, root.val)
        return (False, root.val)