5.1.1 Binary Tree Level Order Traversal

Notes:
	Given a binary tree, return the level order traversal of its nodes' values.
	(ie, from left to right, level by level).
	For example:
	Given binary tree {3,9,20,#,#,15,7},
	3
	/ \
	9 20
	/ \
	15 7
	return its level order traversal as:
	[
	[3],
	[9,20],
	[15,7]
	]
	Solution: 1. Use queue. In order to seperate the levels, use 'NULL' as the end indicator of one level.
	2. DFS.
	*/
	/**
	* Definition for binary tree
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/

vector<vector<int> > levelOrder_1(TreeNode *root) {
        vector<vector<int> > res;
        if (!root) return res;
        queue<TreeNode *> q;
        q.push(root);
        q.push(NULL);
        vector<int> level;
        while (true)
        {
            TreeNode *node = q.front(); q.pop();
            if (!node)
            {
                res.push_back(level);
                level.clear();
                if (q.empty()) break; // end
                q.push(NULL);
            }
            else
            {
                level.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return res;
    }