《剑指offer》-逐层打印二叉树

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;
}