human
I found an interesting problem on https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree/:
One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as #.

     _9_
    /   \
   3     2
  / \   / \
 4   1  #  6
/ \ / \   / \
# # # #   # #

For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.

Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

Each comma separated value in the string must be either an integer or a character '#' representing null pointer.

You may assume that the input format is always valid, for example it could never contain two consecutive commas such as "1,,3".

Example 1:

Input: "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

Input: "1,#"
Output: false

Example 3:

Input: "9,#,#,1"
Output: false

I tried it in Python, but could not do it. Can you solve it?

 Give me a Python program to find an algorithm without reconstructing the tree?
assistant
#Here is Python program to find an algorithm without reconstructing the tree.

class Solution:
     def isValidSerialization(self, preorder):
         """
         :type preorder: str
         :rtype: bool
         """   
         preorder, first = preorder.split(","), preorder.split(",")
         def backward(index):
             if index >= len(preorder) or index < 0: return 
             if index+1<len(preorder) and preorder[index+1] == preorder[index] == "#" and index-1 >= 0 and preorder[index-1] != "#":
                 preorder.pop(index)
                 preorder.pop(index)
                 preorder[index-1] = "#"
                 backward(index-2)
             else: backward(index+1)
         backward(0)
         return True if (preorder != first and preorder == ["#"]) or (preorder == first == ["#"]) else False