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],
  [20,9],
  [15,7]
]

/**
 * 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<vector<int> > zigzagLevelOrder(TreeNode *root) {
        vector<vector<int> > result;
        if (root == NULL)
        {
            return result;
        }

        queue<TreeNode*> buf;
        bool left = true;
        buf.push(root);
        buf.push(NULL);
        vector<int> temp;
        stack<int> tempRight;
        while (!buf.empty())
        {
            TreeNode *front = buf.front();
            if (front != NULL)
            {
                if (left)
                {
                    temp.push_back(front->val);
                }
                else
                {
                    tempRight.push(front->val);
                }

                if (front->left)
                {
                    buf.push(front->left);
                }
                if (front->right)
                {
                    buf.push(front->right);
                }
            }
            else
            {
                if (!left)
                {
                    while (!tempRight.empty())
                    {
                        int top = tempRight.top();
                        temp.push_back(top);
                        tempRight.pop();
                    }
                }
                result.push_back(temp);
                temp.clear();
                if (buf.size() > 1)
                {
                    if (left)
                    {
                        left = false;
                    }
                    else
                    {
                        left = true;
                    }
                    buf.push(NULL);
                }
            }
            buf.pop();
        }

        return result;
    }
};