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
#
class Solution:
    def sumNumbers(self , root: TreeNode) -> int:
        # write code here
        def dfs(root,pre_sum):
            if not root:
                return 0
            pre_sum = pre_sum*10+root.val
            if not root.left and not root.right:
                return pre_sum
            else:
                return dfs(root.left,pre_sum)+dfs(root.right,pre_sum)
        return dfs(root,0)