题解 | #字符串分隔#思路就是用队列存每层所有结点，然后把结点的值存进vector，再根据层数反转vector

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    vector<vector<int> > Print(TreeNode* pRoot) {
        queue<TreeNode*> q;
        vector<int> a;
        vector<vector<int>> v;
        int n;
        TreeNode* t;
        q.push(pRoot);
        int count = 0;
        if(pRoot == nullptr){
            return {};
        }
        while(!q.empty()){
            a.clear();
            n = q.size();
            for(int i = 0;i < n;i++){
                t = q.front();
                a.push_back(t->val);
                q.pop();
                if(t->left){
                    q.push(t->left);
                }
                if(t->right){
                    q.push(t->right);
                }
            }
            if(count%2 == 0){
                v.push_back(a);
            }
            else{
                reverse(a.begin(),a.end());
                v.push_back(a);
            }
            count++;
        }
        return v;
    }
    
};