Binary Tree part 3

https://leetcode.com/problems/count-univalue-subtrees/

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def countUnivalSubtrees(self, root: TreeNode) -> int:
        self.res = 0
        def helper(node):
            if node is None:return True, None
            ok, val = helper(node.left)
            ok1, val1 = helper(node.right)
            if ok and ok1 and (val is None or val==node.val) and (val1 is None or val1==node.val):
                self.res += 1
                return True, node.val
            else:return False, -1
        helper(root)
        return self.res

Find Duplicate Subtrees - LeetCode

In order fails in the following case

[0,0,0,0,null,null,0,null,null,null,0]

class Solution:
    def findDuplicateSubtrees(self, root: TreeNode) -> List[TreeNode]:
        d = collections.defaultdict(list)
        def helper(node):
            if node is None:return [None]
            left = helper(node.left)
            right = helper(node.right)
            # in order fail
            #values = left + [node.val] + right
            # preorder works
            #values = [node.val] +left +  right
            # postorder also works
            values = left +  right + [node.val]
            d[tuple(values)].append(node)
            return values
        helper(root)
        return [d[key][0] for key in d if len(d[key])>=2]

In order doesn’t work for this case
TreeA:
0
/
0

TreeB:
0
\
0
However, if we change left None to “(“, right None to “)”, all traverse will work.
For this example.
Tree A
In order: (0)0)
pre order: 00())
post order: ()0)0
Tree B:
in order: (0(0)
pre order: (0()0
post order: (()00

class Solution:
    def findDuplicateSubtrees(self, root: TreeNode) -> List[TreeNode]:
        d = collections.defaultdict(list)
        def helper(node):
            if node is None:return None
            left = helper(node.left)
            right = helper(node.right)
            if left is None: left = ['(']
            if right is None: right = [')']
            values = left +[node.val]+  right 
            d[tuple(values)].append(node)
            return values
        helper(root)
        return [d[key][0] for key in d if len(d[key])>=2]