二叉树迭代式后序遍历实现逻辑解析：双栈法思路溯源

Great question—this is exactly the kind of deep dive that turns "memorized tricks" into reusable problem-solving intuition. The two-stack approach isn’t just a lucky observation; it’s derived from combining two core ideas: the inverse relationship between traversal orders and the last-in-first-out (LIFO) behavior of stacks.

Let’s walk through the thought process step by step to see how you could derive this method yourself:

1. Start with the core goal of post-order traversal #

Post-order traversal follows the sequence: Left → Right → Root. If you write this sequence backwards, you get Root → Right → Left—this is a "modified pre-order traversal" (standard pre-order is Root → Left → Right). That’s the first critical insight: post-order is just the reverse of a flipped pre-order.

2. Use a stack to implement the modified pre-order #

Stacks are perfect for traversals because their LIFO behavior lets us "delay" processing child nodes. For standard pre-order, we push the right child first, then the left—so when we pop, we process the left child first. To get our modified Root → Right → Left sequence, we just flip that push order:

• Push the root node onto the first stack.
• While the first stack isn’t empty:
  • Pop a node, process it (but instead of outputting it immediately, we’ll store it).
  • Push its left child first, then its right child onto the first stack.
    This way, when we pop from the first stack, we’ll always process the right child before the left—giving us that Root → Right → Left sequence.

3. Use a second stack to reverse the sequence #

Remember, we need the reverse of Root → Right → Left to get post-order. Instead of collecting the modified pre-order results in a list and reversing it, we can use a second stack: every time we pop a node from the first stack, we push it onto the second stack. Since stacks reverse order naturally, popping nodes from the second stack will give us exactly the Left → Right → Root post-order sequence.

Example to make it concrete #

Let’s take a simple tree: Root (A) → Left (B), Right (C)

1. Push A to stack 1.
2. Pop A, push to stack 2. Push B, then C to stack 1.
3. Pop C, push to stack 2. C has no children, so move on.
4. Pop B, push to stack 2. B has no children, so move on.
5. Now pop from stack 2: B → C → A—perfect post-order.

Key thinking directions to derive such methods #

If you want to come up with similar traversal techniques on your own, follow these steps:

• Reverse-engineer the target sequence: Write down the order you need, then see if it’s the reverse (or a simple modification) of a more familiar traversal (like pre-order or in-order).
• Leverage stack behavior: Remember that stacks flip order—use this to your advantage when you need to delay processing nodes or reverse a sequence.
• Break the problem into smaller steps: Instead of trying to get the final sequence directly, build an intermediate sequence that’s easier to generate, then transform it to get your desired result.

内容的提问来源于stack exchange，提问作者anjkhade