Leetcode: Binary Tree Postorder Traversal（二叉树后序遍历）

题目：
Given a binary tree, return the postorder traversal of its nodes’ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

递归解法（C++版本）：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
private:
    vector<int> result;
public:
    vector<int> postorderTraversal(TreeNode *root)
    {
        if (root)
        {
            postorderTraversal(root->left);
            postorderTraversal(root->right);
            result.push_back(root->val);
        }
        return result;
    }
};

二叉树非递归后序遍历思路：
二叉树非递归后续遍历是三种遍历中最难的，因为后续遍历最后访问根节点，操作中不好回溯到根节点，而前序遍历和中序遍历中，每次循环遍历中都是从根节点开始回溯很容易。
网上好些代码都是使用一个栈Stack进行操作的，代码不易懂，读起来很吃力。
其实，可以使用两个Stack来进行二叉树非递归后续遍历。先访问根节点，入栈（A栈），记住根节点，出栈进入另外一个栈（B栈），然后该根节点左右子树入栈（A栈）。最后输出B栈的结果就OK了。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
public:
    vector<int> postorderTraversal(TreeNode *root)
    {
        vector<int> result;
        if (!root) return result;
        stack<TreeNode*> outputStack;
        stack<TreeNode*> middleStack;
        middleStack.push(root);
        TreeNode *node = nullptr;
        while (!middleStack.empty())
        {
            node = middleStack.top();
            middleStack.pop();
            outputStack.push(node);
            if (node->left)
            {
                middleStack.push(node->left);
            }
            if (node->right)
            {
                middleStack.push(node->right);
            }
        }
        while (!outputStack.empty())
        {
            result.push_back(outputStack.top()->val);
            outputStack.pop();
        }
        return result;
    }
};

C#非递归参考代码：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution
{
    public IList<int> PostorderTraversal(TreeNode root)
    {
        IList<int> result = new List<int>();
        if (root == null) return result;
        Stack<TreeNode> outputStack = new Stack<TreeNode>();
        Stack<TreeNode> middleStack = new Stack<TreeNode>();
        middleStack.Push(root);
        TreeNode node = null;
        while (middleStack.Count != 0)
        {
            node = middleStack.Pop();
            outputStack.Push(node);
            if (node.left != null) middleStack.Push(node.left);
            if (node.right != null) middleStack.Push(node.right);
        }
        while (outputStack.Count != 0)
        {
            result.Add(outputStack.Pop().val);
        }
        return result;
    }
}