LeetCode OJ: Binary Tree Postorder Traversal

转战LeetCode，开始不适应，只好先从简单的题目开始

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?

分析：看到类方法的返回值是vector<int>，就知道不能用递归了，循环后续遍历二叉树也是当年考研的经典题目了，下面的代码中使用了两个数组栈，操作起来也蛮方便的

/**
* Definition for binary tree
* 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>* v_ptr = new vector<int>;
vector<int> vec;
if (root == NULL) return vec;
else{
const int MAXSIZE = 10000;
TreeNode* stack[MAXSIZE];
int mark[MAXSIZE];
int cnt = 0;
stack[cnt] = root;
mark[cnt] = 0; //当前节点的左右孩子均为未访问
++cnt;
while (cnt != 0){
switch (mark[cnt-1]){
case 0: //若从未被取出过，则访问左孩子
++(mark[cnt-1]); //标记为取出过一次
if(stack[cnt-1]->left != NULL){
stack[cnt] = stack[cnt-1]->left;
mark[cnt] = 0;
++cnt;
}
break;
case 1: //若被取出过一次，则访问右孩子
++(mark[cnt-1]); //标记为取出过两次
if(stack[cnt-1]->right != NULL){
stack[cnt] = stack[cnt-1]->right;
mark[cnt] = 0;
++cnt;
}
break;
case 2: //若被取出过两次，则访问当前结点，并出栈
vec.push_back(stack[cnt-1]->val);
--cnt;
break;
}
}
return vec;
}
}
};