<p>You are given a positive integer <code>n</code> representing the number of nodes of a <strong>Directed Acyclic Graph</strong> (DAG). The nodes are numbered from <code>0</code> to <code>n - 1</code> (<strong>inclusive</strong>).</p>

<p>You are also given a 2D integer array <code>edges</code>, where <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>]</code> denotes that there is a <strong>unidirectional</strong> edge from <code>from<sub>i</sub></code> to <code>to<sub>i</sub></code> in the graph.</p>

<p>Return <em>a list</em> <code>answer</code><em>, where </em><code>answer[i]</code><em> is the <strong>list of ancestors</strong> of the</em> <code>i<sup>th</sup></code> <em>node, sorted in <strong>ascending order</strong></em>.</p>

<p>A node <code>u</code> is an <strong>ancestor</strong> of another node <code>v</code> if <code>u</code> can reach <code>v</code> via a set of edges.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/12/12/e1.png" style="width: 322px; height: 265px;" />
<pre>
<strong>Input:</strong> n = 8, edgeList = [[0,3],[0,4],[1,3],[2,4],[2,7],[3,5],[3,6],[3,7],[4,6]]
<strong>Output:</strong> [[],[],[],[0,1],[0,2],[0,1,3],[0,1,2,3,4],[0,1,2,3]]
<strong>Explanation:</strong>
The above diagram represents the input graph.
- Nodes 0, 1, and 2 do not have any ancestors.
- Node 3 has two ancestors 0 and 1.
- Node 4 has two ancestors 0 and 2.
- Node 5 has three ancestors 0, 1, and 3.
- Node 6 has five ancestors 0, 1, 2, 3, and 4.
- Node 7 has four ancestors 0, 1, 2, and 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/12/12/e2.png" style="width: 343px; height: 299px;" />
<pre>
<strong>Input:</strong> n = 5, edgeList = [[0,1],[0,2],[0,3],[0,4],[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
<strong>Output:</strong> [[],[0],[0,1],[0,1,2],[0,1,2,3]]
<strong>Explanation:</strong>
The above diagram represents the input graph.
- Node 0 does not have any ancestor.
- Node 1 has one ancestor 0.
- Node 2 has two ancestors 0 and 1.
- Node 3 has three ancestors 0, 1, and 2.
- Node 4 has four ancestors 0, 1, 2, and 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>0 &lt;= edges.length &lt;= min(2000, n * (n - 1) / 2)</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= from<sub>i</sub>, to<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>from<sub>i</sub> != to<sub>i</sub></code></li>
	<li>There are no duplicate edges.</li>
	<li>The graph is <strong>directed</strong> and <strong>acyclic</strong>.</li>
</ul>

class Solution:
    def getAncestors(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        def bfs(i):
            queue = deque()
            queue.append(i)
            visited = set()

            while queue:
                node = queue.popleft()
                for child in adjList[node]:
                    if child not in visited:
                        visited.add(child)
                        queue.append(child)
                        childToAncestors[child].append(i)

        # create adj list
        adjList = {i:[] for i in range(n)}
        for u, v in edges:
            adjList[u].append(v)
        print(adjList)

        childToAncestors = [[] for _ in range(n)]

        # run bfs through each node, geting its ancestors
        for i in range(n):
            bfs(i)

        return childToAncestors
559