LeetCode OJ：Binary Tree Postorder Traversal（后序遍历二叉树）

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?

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递归当然很容易实现拉，但是要求是不使用递归来实现，先贴一个递归的代码：

 class Solution {
 public:
     vector<int> postorderTraversal(TreeNode* root) {
         if(!root) return ret;
         tranverse(root);
         return ret;
     }

     void tranverse(TreeNode * root)
     {
         if(!root) return;
         tranverse(root->left);
         tranverse(root->right);
         ret.push_back(root->val);
     }
 private:
     vector<int> ret;
 };

下面是非递归实现的代码，用一个栈来保存数据：
注意左右子树的压栈顺序

 /**
  * 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) {
         vector<int> ret;
         if(!root) return ret;
         stack<TreeNode *> s;
         s.push(root);
         while(!s.empty()){
             TreeNode * t = s.top();
             if(!t->left && !t->right){
                 ret.push_back(t->val);
                 s.pop();
                 continue;
             }
             if(t->right){
                 s.push(t->right);
                 t->right = NULL;
             }
             if(t->left){
                 s.push(t->left);
                 t->left = NULL;
             }
         }
         return ret;
     }
 };