102/107. Binary Tree Level Order Traversal/II

原文题目：

102. Binary Tree Level Order Traversal

107. Binary Tree Level Order Traversal II

读题：

102. 层序遍历二叉树，每一层作为一个数组，从上到下输出

107.层序遍历二叉树，每一层作为一个数组，反过来从下到上输出

两者只有最后一行的存储方式不一致

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        res = []
        queue = [root]
        while queue:
            level = []
            for i in range(len(queue)):
                node = queue.pop()
                level.append(node.val)
                if node.left:
                    queue.insert(0,node.left)
                if node.right:
                    queue.insert(0,node.right)
            res.append(level) #102
            #res.insert(0,level) #107

　　

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) 
	{
		vector<vector<int>> res;
		if (NULL == root) return res;

		queue <TreeNode*> q;
		q.push(root);
		while(!q.empty())
		{
			vector <int> oneLevel;
			int size  =q.size();
			for (int i = 0; i<size;++i)
			{
				TreeNode *node = q.front();
				q.pop();
				oneLevel.push_back(node->val);
				if (node->left) q.push(node->left);
				if (node->right) q.push(node->right);
			}
			res.push_back(oneLevel);
		}
        //对层次遍历结果进行翻转  
        //reverse(res.begin(), res.end()); //107则增加这一行
		return res;
	}
};