LeetCode OJ--Binary Tree Postorder Traversal

http://oj.leetcode.com/problems/binary-tree-postorder-traversal/

树的后序遍历，可以使用递归，也可以使用栈，下面是栈的实现代码

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

 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> ans;
        if(root == NULL)
            return ans;
        stack<TreeNode*> myStack;
        myStack.push(root);
        TreeNode* lastPop = NULL;
        TreeNode* nodeTop = NULL;
        while(!myStack.empty())
        {
            nodeTop = myStack.top();
            if(nodeTop->right && lastPop != nodeTop->right)
                myStack.push(nodeTop->right);
            else if(nodeTop->right && lastPop == nodeTop->right)
            {
                ans.push_back(nodeTop->val);
                lastPop = nodeTop;
                myStack.pop();
                continue;
            }
            if(nodeTop->left && (nodeTop->right &&lastPop != nodeTop->right || nodeTop->right == NULL && lastPop != nodeTop->left))
                myStack.push(nodeTop->left);
            else if(nodeTop->left && (nodeTop->right && lastPop == nodeTop->right || nodeTop->right == NULL && lastPop == nodeTop->left))
            {
                ans.push_back(nodeTop->val);
                lastPop = nodeTop;
                myStack.pop();
                continue;
            }
            if(nodeTop->left == NULL && nodeTop->right == NULL)
            {
                ans.push_back(nodeTop->val);
                lastPop = nodeTop;
                myStack.pop();
            }
        }
        return ans;
    }
};

int main()
{ 
    TreeNode *root = new TreeNode(1);
    TreeNode *n2 = new TreeNode(2);
    //TreeNode *n3 = new TreeNode(2);
    TreeNode *n4 = new TreeNode(4);
    //TreeNode *n5 = new TreeNode(5);*/
    TreeNode *n6 = new TreeNode(3);
    TreeNode *n7 = new TreeNode(4);

    root->left = n2;
    //root->right = n3;
    n2->left = n4;
    //n2->right = n5;*/
    //n3->left = n6;
    //n6->right = n7;
    Solution myS;
    
    myS.postorderTraversal(root);
    return 0;
}