LeetCode: Binary Tree Inorder Traversal

/**
 * Given a binary tree, return the inorder traversal of its nodes' values.

 * For example:
 * Given binary tree [1,null,2,3],
 * return [1,3,2]
*/

/// Test Unit
/// {1,2,3,#,#,4,#,#,5}
/* 
 *          1
 *         / \
 *        2   3
 *           /
 *          4
 *           \
 *            5
 */



#include <iostream>
#include <vector>
#include <stack>
using namespace std;


struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

// Recursion
class Solution1 {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> res;
        inorder(root, res);
        return res;
    }
    void inorder(TreeNode *root, vector<int> &res) {
        if (!root) return;
        if (root->left) inorder(root->left, res);
        res.push_back(root->val);
        if (root->right) inorder(root->right, res);
    }
};

//Non-Recursion
class Solution2 {
public:
    vector<int> inorderTraversal(TreeNode *root){
        vector<int> res;
        stack<TreeNode*> s;
        TreeNode *p=root;
        while(p||!s.empty()){
            while(p){
                s.push(p);
                p=p->left;
            }

            p=s.top();          
            res.push_back(p->val);  
            s.pop();
            p=p->right;

        }
        return res;
    }
};




int main(int argc, char** argv) {
  TreeNode root(1);
  TreeNode node2(2);
  TreeNode node3(3);
  TreeNode node4(4);
  TreeNode node5(5);

  root.left = &node2;
  root.right = &node3;
  node3.left = &node4;
  node4.right = &node5;

  Solution2 t;
  vector<int> v = t.inorderTraversal(&root);     

  /*for (auto i = v.begin(): i != v.end();++i)  //c++11
     cout << *i << " ";
  cout << endl; 
  */

  for(auto i:v)
    cout<< i <<" ";
  cout<<endl;


  return 0;
}