题解 | #二叉树根节点到叶子节点的所有路径和#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root TreeNode类 
# @return int整型
def dfs(node, pre_sum):
    if node == None:
        return 0
    total_sum = pre_sum * 10 + node.val
    if node.right == None and node.left == None:
        return total_sum
    else:
        return dfs(node.left, total_sum) + dfs(node.right, total_sum)

class Solution:
    def sumNumbers(self , root: TreeNode) -> int:
        # write code here
        pre_sum = 0
        total_sum = dfs(root, pre_sum)
        return total_sum