和上一个题目一样，我只是加入了一个flag 值用来标记该行是否时逆向输出

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool flag;
    vector<vector<int>> a;
    vector<int> b;
    vector<TreeNode*> reserve;
    queue<TreeNode*> q;
    vector<vector<int>> printFromTopToBottom(TreeNode* root) {
        if(!root)   return a;

        reserve.push_back(root);
        while(reserve.size()){
            //将一行的节点全部压入队列
            for(auto item : reserve){
                q.push(item);
            }
            //将存储那一行的节点全部清除
            while(reserve.size())   reserve.pop_back();
            //将队列里面的元素进行遍历
            while(q.size()){
                TreeNode* t = q.front();
                q.pop();
                //将这一行的所有元素存储进一维数组
                b.push_back(t -> val);
                //将这一行的所有的子节点全部存储进reserve
                if(t -> left != NULL)   reserve.push_back(t -> left);
                if(t -> right != NULL)  reserve.push_back(t -> right);
            }
            //将一行的数据存储进二维数组
            if(!flag)   flag = true;
            else{
                flag = false;
                reverse(b.begin(), b.end());
            }
            a.push_back(b);
            //将b里面的元素清空，为下一行做准备
            while(b.size()) b.pop_back();
        }
        return a;
    }
};