链接
https://leetcode.cn/problems/sum-root-to-leaf-numbers/description/
思路
时刻记住，DFS是递归的一种。而解决递归，最朴素的思路就是：递归的定义就是递归的解。
题目要求我们求根节点到叶子结点的和，我们要提供一个值保持其状态，退出条件直接用题目给的限定即可。
代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def sumNumbers(self, root) -> int:
res = 0
def dfs(node, cur):
nonlocal res
if node is None:
return
if node.left is None and node.right is None:
res += cur * 10 + node.val
dfs(node.left, cur * 10 + node.val)
dfs(node.right, cur * 10 + node.val)
dfs(root, 0)
return res
标签：node,None,right,val,self,DFS,129,求根,left From： https://www.cnblogs.com/bjfu-vth/p/17766961.html