5.1.2 Binary Tree Level Order Traversal II

Notes:
	Given a binary tree, return the bottom-up level order traversal of its nodes' values.
	(ie, from left to right, level by level from leaf to root).
	For example:
	Given binary tree {3,9,20,#,#,15,7},
	3
	/ \
	9 20
	/ \
	15 7
	return its bottom-up level order traversal as:
	[
	[15,7]
	[9,20],
	[3],
	]
	Solution: Queue version. On the basis of 'Binary Tree Level Order Traversal', reverse the final vector.
	*/
	/**
	* 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> > levelOrderBottom(TreeNode *root) {
        vector<vector<int>> root2leaf;
        queue<TreeNode *> q;
        if (!root) return root2leaf;
        q.push(root);
        q.push(NULL);   // end indicator of one level
        vector<int> level;
        while (true)
        {
            TreeNode *node = q.front(); q.pop();
            if (node)
            {
                level.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            else
            {
                root2leaf.push_back(level);
                level.clear();
                if (q.empty()) break;    // CAUTIOUS! infinite loop
                q.push(NULL);
            }
        }
    	// reverse
        reverse(root2leaf.begin(), root2leaf.end());
        return root2leaf;
    }
};