题目：
Given a list, rotate the list to the right by k places, where k is non-negative.
For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.
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?
分类：Tree Stack
代码：二叉树非递归后序遍历
/**
* 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) {
stack<TreeNode*> s;
vector<int> nodes;
TreeNode* cur = root;//u当前访问的结点
TreeNode* lastNode = NULL;//上次访问的结点
while(cur || !s.empty())
{
//一直向左走直到为空为止
while(cur)
{
s.push(cur);
cur = cur->left;
}
cur = s.top();
//如果结点右子树为空或已经访问过，访问当前结点
if(cur->right == NULL || cur->right == lastNode)
{
nodes.push_back(cur->val);
lastNode = cur;
s.pop();
cur = NULL;
}
else
cur = cur->right;
}
return nodes;
}
};