题目描述
从上到下按层打印二叉树，同一层结点从左至右输出。每一层输出一行。

乍一看就是一个BFS，但是因为太久没刷题都忘记了要使用queue来作为空间存储容器了。

先参考milolip的代码，写出这样的solution：

class Solution {
public:vector<vector<int> > Print(TreeNode* pRoot) {vector<vector<int> > res;if(pRoot==NULL){return res;}queue<TreeNode*> Q;Q.push(pRoot);Q.push(NULL);vector<int> v;v.push_back(pRoot->val);res.push_back(v);v.clear();while (!Q.empty()){TreeNode* node = Q.front();Q.pop();if (node != NULL){//v.push_back(node->val);//cout << node->val << ends;if (node->left){Q.push(node->left);v.push_back(node->left->val);}if (node->right){Q.push(node->right);v.push_back(node->right->val);}}else if (!Q.empty()){//cout << "test " << endl;Q.push(NULL);res.push_back(v);v.clear();//cout << endl;}}return res;}
};

上面的代码并不太简洁的样子。

另一种写法是从评论区copy来的，又简洁，又非常直观清晰。两层while的嵌套，正好对应到数的层次遍历以及层内逐点遍历。而这种双层嵌套的循环其实并没有增加复杂度，和原来的复杂度是一样的。

class Solution_11 {
public:vector<vector<int> > Print(TreeNode* pRoot) {vector<vector<int> > res;if (pRoot == NULL){return res;}queue<TreeNode*> q;q.push(pRoot);while (!q.empty()){int lo = 0, hi = q.size();vector<int> v;while (lo++ < hi){TreeNode *t = q.front();q.pop();v.push_back(t->val);if (t->left){q.push(t->left);}if (t->right){q.push(t->right);}}res.push_back(v);}return res;}
};

测试代码；

void main_solution_11(){Solution_11 s = Solution_11();TreeNode* a = new TreeNode(8);TreeNode* b1 = new TreeNode(6);TreeNode* b2 = new TreeNode(10);TreeNode* c1 = new TreeNode(5);TreeNode* c2 = new TreeNode(7);TreeNode* c3 = new TreeNode(9);TreeNode* c4 = new TreeNode(1);a->left = b1;a->right = b2;b1->left = c1;b1->right = c2;b2->left = c3;b2->right = c4;vector<vector<int> > q = s.Print(a);for (int i = 0; i < q.size(); i++){for (int j = 0; j < q[i].size(); j++){if (j > 0){cout << " ";}cout << q[i][j];}cout << endl;}
}int main(){main_solution_11();return 0;
}