Leetcode[tree]: Binary Tree Zigzag level Order traversal

given a binary tree, return the zigzag level order traversal of its nodes ' values. (ie, from left-to-right, then right-to-left for the next level and alternate between).
for example:
given binary Tree {3,9,20,#,#,15,7},
3
/\
9 20
15 7
return its zigzag level order traversal as:
[
[3],< BR style= "" >[20,9],
[15,7]
"

My iterative solution is as follows:

vector<vector<int> > Zigzaglevelorder (TreeNode *root) {    vector<vector<int> > res;    Vector<treenode *> iter;    if (root) iter.push_back (root);    int isforward = 1;    while (!iter.empty ()) {        int size = Iter.size ();        Vector<int> level;        for (int i = 0; i < size; ++i) {            level.push_back (iter[i]->val);            if (iter[i]->left) iter.push_back (iter[i]->left);            if (iter[i]->right) iter.push_back (iter[i]->right);        }        if (!isforward) reverse (Level.begin (), Level.end ());        Res.push_back (level);        Iter.erase (Iter.begin (), Iter.begin () + size);        Isforward ^= 0x01;    return res;}

My recursive solution is as follows:

Class Solution {public:    vector<vector<int> > Zigzaglevelorder (TreeNode *root) {        res.clear ();        if (!root) return res;        Zigzaglevelorder (root, 1);        return res;    } Private:    vector<vector<int> > Res;    void Zigzaglevelorder (TreeNode *root, int level) {        if (res.size () < level) res.resize (level);        if (level% 2) res[level-1].push_back (root->val);        Else Res[level-1].insert (Res[level-1].begin (), root->val);        if (root->left) Zigzaglevelorder (Root->left, level + 1);        if (root->right) Zigzaglevelorder (root->right, level + 1);    }};

The time performance of the two algorithms is the same as shown in:

Leetcode[tree]: Binary Tree Zigzag level Order traversal