All Paths from Source Lead to Destination

lcpy gen -n 1059   # by problem number
lcpy gen -s all_paths_from_source_lead_to_destination   # by problem name

Input: n = 3, edges = [[0,1],[0,2]], source = 0, destination = 2
Output: false
Explanation: It is possible to reach and get stuck on both node 1 and node 2.

Input: n = 4, edges = [[0,1],[0,3],[1,2],[2,1]], source = 0, destination = 3
Output: false
Explanation: We have two possibilities: to end at node 3, or to loop over node 1 and node 2 indefinitely.

Input: n = 4, edges = [[0,1],[0,2],[1,3],[2,3]], source = 0, destination = 3
Output: true

class Solution:
    # Time: O(n + e)
    # Space: O(n + e)
    def leads_to_destination(
        self, n: int, edges: list[list[int]], source: int, destination: int
    ) -> bool:
        graph: list[list[int]] = [[] for _ in range(n)]
        for a, b in edges:
            graph[a].append(b)
        if graph[destination]:
            return False

        state = [0] * n

        def dfs(i: int) -> bool:
            if state[i]:
                return state[i] == 2
            if not graph[i]:
                return i == destination
            state[i] = 1
            for j in graph[i]:
                if not dfs(j):
                    return False
            state[i] = 2
            return True

        return dfs(source)