Hey there! It makes total sense to feel confused even after checking examples and videos—topological sort can feel counterintuitive at first. Let’s unpack your question and clarify how things actually work.

First, let’s address your core question: your initial understanding has a kernel of truth, but it’s missing some key details tied to the actual purpose of topological sort.

What Topological Sort Actually Focuses On #

Topological sort is exclusively for Directed Acyclic Graphs (DAGs). Its goal is to produce a sequence of nodes where for every directed edge u → v, node u appears before node v in the sequence. In short, it’s all about respecting dependencies (e.g., "you must finish task A before starting task B").

How Your Idea Aligns (and Diverges) with Common Algorithms #

You mentioned starting with visited/unvisited queues and recording nodes after all their children are processed—this is very close to the DFS-based topological sort method, but with a critical twist:

• DFS Post-Order + Reverse: When using DFS for topological sort, you traverse a node’s all children first. Once all children are fully processed (not just visited), you add the current node to a list. However, this list will be in reverse topological order! You need to reverse it at the end to get the valid sequence.

  For example:

  • Suppose we have a graph A → B, A → C, B → D
  • DFS traversal would process D first (no children), add to list → [D]
  • Then process B (all children done), add to list → [D, B]
  • Then process C (no children), add to list → [D, B, C]
  • Finally process A (all children done), add to list → [D, B, C, A]
  • Reverse the list to get the valid topological order: [A, C, B, D]

• Kahn’s Algorithm (Queue-Based, No Reversal): This is another common method that uses a queue, but it’s focused on in-degrees (number of incoming edges) rather than visited/unvisited states:

  1. Start by enqueueing all nodes with an in-degree of 0 (no dependencies).
  2. While the queue isn’t empty:
     • Dequeue a node, add it to the result sequence.
     • For each of its neighbors, decrement their in-degree by 1. If any neighbor’s in-degree becomes 0, enqueue it.
  3. If the result sequence has fewer nodes than the graph, the graph has a cycle (no valid topological sort exists).

Key Corrections to Your Initial Idea #

• It’s not just about "visiting all children"—it’s about ensuring that all dependencies of a node are resolved before the node appears in the sequence.
• The "visited/unvisited" approach works only if you account for post-order traversal and reverse the final list (for the DFS method).
• Topological sort isn’t a single fixed sequence—many valid sequences can exist for a single DAG, as long as dependencies are respected.

Quick Recap #

Your intuition about processing children first is on the right track, but remember:

1. Topological sort is about dependencies, not just traversal order.
2. The DFS method requires reversing the post-order list to get the correct sequence.
3. Kahn’s algorithm offers an alternative queue-based approach that doesn’t need reversal.

内容的提问来源于stack exchange，提问作者M. Stephens