[leetcode] Binary Tree Postorder Traversal

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].

给定一棵二叉树，使用非递归的方法返回后序遍历结果。后序遍历需要先访问左子树，然后再访问右子树，最后访问自己。因此需要记录该节点的左右子树是否已经全部访问，如果访问完毕，那么就可以访问自己了。

代码如下：

 1 class Solution {

 2 public:

 3     vector<int> postorderTraversal(TreeNode *root) {

 4         // IMPORTANT: Please reset any member data you declared, as

 5         // the same Solution instance will be reused for each test case.

 6         vector<int> ans;

 7         list<TreeNode*> node_list;

 8         if(root == NULL) return ans;

 9         node_list.push_front(root);

10         TreeNode *head = root;

11         while(!node_list.empty())

12         {

13             TreeNode *cur = node_list.front();

14             if(cur -> right == head || cur -> left == head || ((cur -> right == NULL) && (cur -> left == NULL)))

15             {

16                 node_list.pop_front();

17                 ans.push_back(cur -> val);

18                 head = cur;

19             }

20             else

21             {

22                 if(cur -> right != NULL) node_list.push_front(cur -> right);

23                 if(cur -> left != NULL) node_list.push_front(cur -> left);

24             }

25         }

26     }

27 };