如何在Python递归期间使用共享变量
How to use a shared variable during Recursion in Python

您的变量min_path的作用域为main方法，不适用于属于您的类对象的函数。 为了在backtrack方法中引用它，您要么需要

1. 将min_path定义为backtrack方法中的局部变量（由于递归调用它可能会引起其他问题）：

class Solution:
def minPathSum(self, grid: List[List[int]]) -> int:
    if not grid:
        return

    R = len(grid)
    C = len(grid)
    min_path_main = None

    min_path_main = backtrack(grid, 0, 0,0)

    return min_path_main

  def backtrack(grid, i, j, current_path):
    min_path = None
    if i >= R or j>= C:
        break
      current_path += grid[i][j]
      if i == R-1 and j == C-1:
        if not min_path:
            min_path = current_path
        else:
            min_path = min(min_path, current_path)
            print(min_path)
            break
    # it's possible you'll need extra logic here to manage the output. 
    output1 = backtrack(grid, i, j+1, current_path)
    output2 = backtrack(grid, i+1, j, current_path)

  return min_path

2. 将变量定义为类变量：

class Solution:
def __init__(self):
  self.min_path = None

def minPathSum(self, grid: List[List[int]]) -> int:
    if not grid:
        return

    R = len(grid)
    C = len(grid)

    backtrack(grid, 0, 0,0)

    return self.min_path

def backtrack(grid, i, j, current_path):
    if i >= R or j>= C:
        break
      current_path += grid[i][j]
      if i == R-1 and j == C-1:
        if not self.min_path:
            self.min_path = current_path
        else:
            self.min_path = min(self.min_path, current_path)
            print(self.min_path)
            break
    backtrack(grid, i, j+1, current_path)
    backtrack(grid, i+1, j, current_path)

  return

可能需要玩一点。 我没有输入数据，所以没有运行代码。 但是这些就是您实施我提供的解决方案的方式。

由于目标是简单地找到最小和，因此无需跟踪路径的坐标。 相反，您可以从要执行的步骤的角度使用网格剩余的内容进行递归调用：

def min_sum(g):
    if not g or not g[0]:
        return float('inf')
    return g[0][0] + ((len(g) > 1 or len(g[0]) > 1) and min(map(min_sum, (g[1:], [*zip(*g)][1:]))))

以便：

min_sum([
  [1,3,1],
  [1,5,1],
  [4,2,1]
])

返回： 7