70. Climbing Stairs python

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

这是一个简单的爬楼梯的题目，可以一次爬一层，也可以爬两层。如果爬一层就是i-1的情况，如果爬两层就是i-2的情况，所以

l[i]=l[i-1]+l[i-2]

class Solution:
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """      
        l = [1 for i in range(n+1)]      
        for i in range(2,n+1):
            l[i]=l[i-1]+l[i-2]
        return l[n]
    

开始的时候我的写法是

class Solution:
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n==2:
            return 2
        if n==3:
            return 3
        return self.climbStairs(n-1)+self.climbStairs(n-2)

这样字节测试时通过的，但是放到leetcode里面显示递归超过了最大深度，应该是递归的深度太深，而且大量节点重复计算。