LeetCode386 字典序排数

LeetCode386 字典序排数

构造字典树，dfs遍历记录

class Solution:
    def lexicalOrder(self, n: int) -> List[int]:

        ans = []

        def dfs(cur):

            if cur <= n:
                ans.append(cur)
            else:
                return
            for i in range(10):
                dfs(cur * 10 + i)

        for i in range(1, 10): dfs(i)
        return ans