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].

/**
 * Definition for a binary tree node.
 * 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> path;
        postOrder(root, path);
        return path;
    }
    /*
    void postOrder(TreeNode* root, vector<int> &path)
    {
        //递归写法
        if (root)
        {
            postOrder(root->left, path);
            postOrder(root->right, path);
            path.push_back(root->val);
        }
    }
    */
    void postOrder(TreeNode* root, vector<int> &path)
    {
        //非递归写法
        stack<TreeNode *> TreeNodeStack;
        TreeNode *plastvisit = NULL; //记录结点是否访问过
        while (root != NULL || !TreeNodeStack.empty())
        {
            while (root != NULL)
            {
                TreeNodeStack.push(root);
                root = root->left;
            }
            root = TreeNodeStack.top();
            if (root->right == NULL || plastvisit == root->right)
            {
                path.push_back(root->val);
                plastvisit = root;
                TreeNodeStack.pop();
                root = NULL;
            }
            else
                root = root->right;
        }
    }
};