All Ancestors of a Node in a Directed Acyclic Graph
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2100-2199/2192.All%20Ancestors%20of%20a%20Node%20in%20a%20Directed%20Acyclic%20Graph/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2100-2199/2192.All%20Ancestors%20of%20a%20Node%20in%20a%20Directed%20Acyclic%20Graph/images/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>
Depth-First Search; Breadth-First Search; Graph; Topological Sort
public class Solution {
    private int n;
    private List<int>[] g;
    private IList<IList<int>> ans;

    public IList<IList<int>> GetAncestors(int n, int[][] edges) {
        g = new List<int>[n];
        this.n = n;
        for (int i = 0; i < n; i++) {
            g[i] = new List<int>();
        }
        foreach (var e in edges) {
            g[e[0]].Add(e[1]);
        }
        ans = new List<IList<int>>();
        for (int i = 0; i < n; ++i) {
            ans.Add(new List<int>());
        }
        for (int i = 0; i < n; ++i) {
            BFS(i);
        }
        return ans;
    }

    private void BFS(int s) {
        Queue<int> q = new Queue<int>();
        q.Enqueue(s);
        bool[] vis = new bool[n];
        vis[s] = true;
        while (q.Count > 0) {
            int i = q.Dequeue();
            foreach (int j in g[i]) {
                if (!vis[j]) {
                    vis[j] = true;
                    q.Enqueue(j);
                    ans[j].Add(s);
                }
            }
        }
    }
}