Binary Tree Inorder Traversal, iterative solution

inorder_traversal_iterative.cpp

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

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?
*/

/**
 * 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> inorderTraversal(TreeNode *root) {
        //for root
        //      push rightChild to stack
        //      push root (indicated that it's visited) to stack
        //      push leftChild  to stack
        // then while !stack.empty
        //      if topEle visited before, output its value 
        //      else, push its right, itself(mark visited), its left
        
        // How does it work:
        //  if an element was popped and visited, it must be the root  of some subtree
        //        since its left children tree was popped out already, then this root should be output
        //
        stack<TreeNode *> myS;
        unordered_set<TreeNode *> visited;
        vector<int> vInt;
        if (!root)
            return vInt;
        if (root->right)
            myS.push(root->right);
        myS.push(root);    visited.insert(root);
        if (root->left)
            myS.push(root->left);
        
        //working with stack
        while (!myS.empty()) {
            TreeNode *topEle = myS.top(); myS.pop();
            //if previous pushed as root, output data
            if (visited.find(topEle) != visited.end()) {
                vInt.push_back(topEle->val);
                visited.erase(topEle);  //not be used again, save memory
            }
            else {  //not pushed as root, now push right, itself(marked), left
                if (topEle->right)
                    myS.push(topEle->right);
                myS.push(topEle);    visited.insert(topEle);
                if (topEle->left)
                    myS.push(topEle->left);
            }
        }
        return vInt;
    }
    
};