题解 | #二叉树根节点到叶子节点的所有路径和#

helper函数：给定当前结点root和当前的路径和s，返回以root为根节点的树的路径和。

class Solution:
    def sumNumbers(self , root ):
        # write code here
        def helper(root, s):
            if not root:
                return 0
            s = 10*s + root.val
            if not root.left and not root.right:
                return s
            return helper(root.left, s) + helper(root.right, s)
        return helper(root, 0)