scope of variable within a function in python

Question

While trying to solve a problem on leetcode, I ran into an error "local variable referenced before assignment". Below I have given the entire code (the logic of the code itself is not my question). I have defined the variable within the outer function. I read somewhere, that in python, the scope of a variable works in this way - the variable is first searched locally, then in outer functions (if there are any), then global. In my code, no local variable "total" exists, but there is one defined in the outer function. Is my understanding of scope of variables wrong? Also, I used something similar in another problem but instead of an integer, I was using a list, which was similarly defined in the outer function only, and being appended in the inner function. In this case, no such error occurred. What I am doing wrong here? Any clarification is greatly appreciated.

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> int:
        total = 0
        if root is None:
            return 0

        def helper(root, sum, rem):
            if root is None:
                return 


            if root.val == rem:
                total += 1

            if root.left is not None:
                helper(root.left, sum, sum - root.val)    

            if root.right is not None:
                helper(root.right, sum, sum - root.val)

            return 

        helper(root, sum, sum)

        return total
'''

Does this answer your question? [Python global variable/scope confusion](https://stackoverflow.com/questions/30439772/python-global-variable-scope-confusion) — Tomerikoo, Jan 19 '20 at 17:55
Yes it does. Thank you. Can you also let me know is there a better way to approach this kind of problem? Should I send target as an argument as well to the helper function or something else? — Ayush Bisht, Jan 19 '20 at 18:02

score 1 · Accepted Answer · edited Jan 19 '20 at 18:00

To fix this, use a nonlocal declaration:

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> int:
        total = 0
        if root is None:
            return 0

        def helper(root, sum, rem):
            nonlocal total
            if root is None:
                return 


            if root.val == rem:
                total += 1

            if root.left is not None:
                helper(root.left, sum, sum - root.val)    

            if root.right is not None:
                helper(root.right, sum, sum - root.val)

            return 

        helper(root, sum, sum)

        return total

Essentially, total += 1 implicitly tells Python that total is a local variable.

Have a look at this answer

While this might fix the initial problem, it seems like a bad pattern in general. The better solution would be for the inner `helper` function to maintain a purely local count and return that as the result of the function. Then the outer function would use: `total += helper(root, sum, sum)` — jarmod, Jan 19 '20 at 18:02

scope of variable within a function in python

1 Answers1