[LeetCode145]Binary Tree Postorder Traversal

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;
    }
};