I found an interesting problem on https://leetcode.com/problems/sum-root-to-leaf-numbers/:
Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

Note: A leaf is a node with no children.

Example:


Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:


Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

I tried it in Python, but could not do it. Can you solve it?
# Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None
 from copy import deepcopy
 
 
 class Solution:
     def sumNumbers(self, root):
         """
         :type root: TreeNode
         :rtype: int
         """
         self.treeSum = 0
         self.tree(root, [])
         return self.treeSum
     def tree(self, node, path):
         if not node:
             return 
         
 
         path.append(str(node.val))
         
         self.tree(node.left, copy.deepcopy(path))
         self.tree(node.right, copy.deepcopy(path))
         
         if not node.left and not node.right:
             self.treeSum += int(''.join(path))
     
 '''
 class Solution:
     def sumNumbers(self, root):
         self.res = 0
         self.dfs(root, 0)
         return self.res
     
     def dfs(self, root, value):
         if root:
             self.dfs(root.left, value*10+root.val)
             self.dfs(root.right, value*10+root.val)
             
             if not root.left and not root.right:
                 self.res += value*10 + root.val
 '''