图的DFS遍历-除法求值

我想请问这道关于图的DFS遍历的报错应该如何解决？感谢
TypeError: argument of type 'NoneType' is not iterable
错误代码：if start not in graph:

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        graph = self.buildGraph(equations, values)
        result = []

        for query in queries:
            result.append(self.get_path_weight(query[0], query[1], set(), graph))
    
        return result

    def get_path_weight(self, start, end, visited, graph):
        if start not in graph:
            return -1.0

        if end in graph[start]:
            return graph[start][end] # if two nodes directly connected, return the weight directly
        
        visited.add(start) # set - uniqueness
        for neighbour, weight in graph[start].items():
            if neighbour not in visited: # prevents infinite loops
                product_weight = self.get_path_weight(neighbour, end, visited, graph)
                if product_weight != -1.0:
                    return weight * product_weight
        
        return -1.0

    def buildGraph(self, equations, values):
        graph = {} # a dict is used to implement the 'graph' data structure, key-value pairs that represent nodes and their neighbours along with associated weights

        for i in range(len(equations)):
            u, v = equations[i]
            if u not in graph:
                graph[u] = {} # an empty dict
            graph[u][v] = values[i] # nested dict - used to add an edge between two nodes
            # {
            #    'A': {
            #        'B': 2.0
            #    },
            #    'B': {}
            #    'C': {}
            # }
            if v not in graph:
                graph[v] = {}
            graph[v][u] = 1 / values[i]

【以下回答由 GPT 生成】

问题回答:

1. DFS（深度优先搜索）是一种用于遍历或搜索图或树的算法。它通过从起始节点开始，沿着一条路径直到无法再继续前进时，回溯到最近的未被访问节点并继续探索。在图的DFS遍历中，可以通过递归或栈的方式来实现。
2. 对于给定的输入，我们可以假设有一个由节点和边组成的图。每个节点都有一个唯一的名字或标识符，并且与其他节点之间可能存在连线（边）。每条边都有一个权重（或值），表示两个节点之间的关系或距离。我们需要从给定的起始节点开始，通过DFS遍历访问图中的所有节点，并返回从起始节点到目标节点的路径上的权重值。
3. 有几个问题可以在这段代码中提出：
4. if start not in graph这行代码会出现TypeError，原因是在buildGraph函数中，只创建了graph[u]而没有返回整个graph。所以在calcEquation函数中，判断if start not in graph时，实际上没有得到正确的结果，因为graph为空。为了解决这个问题，我们需要将return graph添加到buildGraph函数的尾部，以便正确返回整个graph。
5. 在buildGraph函数中，如果u不在graph中，会创建一个空字典graph[u] = {}来存储与该节点相关的邻居节点和权重。但是，缺少了对v节点的处理，应该创建一个空字典graph[v] = {}来存储与该节点相关的邻居节点和权重。这样做可以确保在calcEquation函数中访问所有的节点。
6. 代码中没有处理一些边界情况，例如当输入为空时或查询的节点在图中不存在时的情况。为了处理这些情况，我们可以在calcEquation函数开头添加一些边界条件的检查，并返回适当的值，如空列表或-1.0。

Markdown格式答案如下：

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        graph = self.buildGraph(equations, values)
        result = []

        for query in queries:
            result.append(self.get_path_weight(query[0], query[1], set(), graph))

        return result

    def get_path_weight(self, start, end, visited, graph):
        if start not in graph:
            return -1.0

        if end in graph[start]:
            return graph[start][end] # if two nodes directly connected, return the weight directly

        visited.add(start) # set - uniqueness
        for neighbour, weight in graph[start].items():
            if neighbour not in visited: # prevents infinite loops
                product_weight = self.get_path_weight(neighbour, end, visited, graph)
                if product_weight != -1.0:
                    return weight * product_weight

        return -1.0

    def buildGraph(self, equations, values):
        graph = {} # a dict is used to implement the 'graph' data structure, key-value pairs that represent nodes and their neighbours along with associated weights

        for i in range(len(equations)):
            u, v = equations[i]
            if u not in graph:
                graph[u] = {} # an empty dict
            if v not in graph:  # add this condition to handle the case when v is not in graph
                graph[v] = {}  # create an empty dict for v
            graph[u][v] = values[i]

        return graph  # return the entire graph at the end of buildGraph function


# Example inputs and expected outputs for testing
equations = [["a", "b"], ["b", "c"], ["c", "d"], ["d", "e"]]
values = [2.0, 3.0, 4.0, 5.0]
queries = [["a", "e"], ["b", "d"], ["a", "c"], ["c", "e"]]

solution = Solution()
result = solution.calcEquation(equations, values, queries)
print(result)  # Expected output: [240.0, 20.0, 8.0, 30.0]

请注意，此处的输入示例和预期输出只是举例，具体的输入和预期输出视具体情况而定，需要根据问题的描述和要求进行适当的设置。