Binary Tree Postorder Traversal <leetcode>

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?

思路：1、非递归遍历  2、递归遍历

非递归：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 
 struct node{
     TreeNode *root;
     bool isFirst;
 };
 
class Solution {
public:
    vector<int>  result;
    vector<int> postorderTraversal(TreeNode *root) {
             vector<node>  str;
             if(root==NULL)  return result;
             TreeNode *p=root;
             while(p!=NULL||str.size()!=0)
             {
                 while(p!=NULL)
                 {
                     node n;
                     n.isFirst=true;
                     n.root=p;
                     str.push_back(n);
                     p=p->left;
                 }
                 if(str.size()!=0)
                 {
                     node temp=str.back();
                     str.pop_back();

                     if(temp.isFirst==true)
                     {
                         temp.isFirst=false;
                         str.push_back(temp);
                         p=temp.root->right;
                     }
                     else
                     {
                         result.push_back(temp.root->val);
                         p=NULL;
                     }
                     
                 }
             }
             return result;
    }
};

2：递归

/**
 * 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>  result;
    vector<int> postorderTraversal(TreeNode *root) {
        digui(root);
        return result;
    }
    
    void digui(TreeNode *root)
    {
        if(root!=NULL)
        {
            digui(root->left);
            digui(root->right);
            result.push_back(root->val);
        }
    }
};