Variable Scope for Nested Function in Python

Question

I'm not understanding why snippet 1 works while the 2nd one doesn't. The 2nd code returns local variable 'ans' referenced before assignment error. Even when I do change 2 to include print(ans) right after def dfs(node): (basically the same structure as 1), it returns the same error. Would appreciate your answer as to why this is happening!

Snippet 1:

def myfunc():
  x = 300
  def myinnerfunc():
    print(x)
  myinnerfunc()

myfunc()

Snippet 2:

from collections import defaultdict

#class Solution:
def minReorder( n, connections):
    seen = set()
    ans = 0 # to count # of changes

    # graph preprocessing 
    cities = defaultdict(list)
    for i in connections:
        a, b = i
        cities[a].append(b)
    
    # dfs
    def dfs(node):
      #print(ans) This is the same structure as snippet 1 but also gives the same error
      #nonlocal ans This fixes the issue!
      for n in cities[node]:
        if n not in seen:
          seen.add(n)
          ans += 1
          dfs(n)


    for i in range(n):
      if i not in seen:
        seen.add(i)
        dfs(i)
        
    print(f'outside {ans}')
    return ans

Test case for snippet 2 if needed:

minReorder(6, [[0,1],[1,3],[2,3],[4,0],[4,5]])

I fixed the error by using nonlocal ans but not sure why it's required for 2 and not for 1