二叉树层序遍历

用队列存储，一般存储的是节点的指针。

1、基础版本：
https://blog.csdn.net/FX677588/article/details/74276513

#include<iostream>
#include<queue>
using namespace std;

struct node{
	int val;
	struct node*left;
	struct node*right;
	node(int v):val(v),left(nullptr),right(nullptr){
	};
};


void Level1(struct node *p)
{
	queue<node*>q;
	q.push(p);
	while (!q.empty()){
		node *tmp = q.front();
		q.pop();
		cout<<tmp->val<<" ";
		if (tmp->left)q.push(tmp->left);
		if (tmp->right)q.push(tmp->right);
	}
}



int main()
{
	
	struct node *BT[6];
	BT[0] = new node(0);
	BT[1] = new node(1);
	BT[2] = new node(2);
	BT[3] = new node(3);
	BT[4] = new node(4);
	//BT[5] = new node(5);
	
	//BT[0]->left = BT[1]; BT[0]->right = BT[2]; BT[1]->left = BT[3]; BT[1]->right = BT[4]; BT[2]->left = BT[5];
	
	BT[0]->left = BT[1]; BT[0]->right = BT[2]; BT[2]->left = BT[3]; BT[2]->right = BT[4];//BT[3]->right = BT[4];
	
	Level1(BT[0]);
	cout<<endl;
	
	
	return 0;
}

2、leetcode:https://leetcode.com/problems/binary-tree-level-order-traversal/
将结果存在二维数组中。需要记录每一层的的数量结束标记。

vector<vector<int> > Level2(struct node *p)
{
	vector<vector<int> >res;
	if (p==nullptr)
	{
		return res;
	}
	
	queue<node*>q;
	q.push(p);
	int count = 1;
	vector<int>level;
	
	while(!q.empty())
	{
		struct node * tmp = q.front();
		q.pop();
		int t = tmp->val;
		level.push_back(t);
		
		if (tmp->left)q.push(tmp->left);
		if (tmp->right)q.push(tmp->right);
		
		count--;
		
		if (count==0)
		{
			count = q.size();
			res.push_back(level);
			level.clear();
		}
		
	}
	return res;
	
}

3. 在（1）的基础上改进，让输出的形式中带有NULL。

void Level3(struct node *p)
{
	if (p==nullptr)return;
	queue<node*>q;
	
	q.push(p);
	int count = 1;
	int cnt_null = 0;
	int tmp_cnt = 1;
	
	while (!q.empty()){

		node * tmp = q.front();
		q.pop();
		if (tmp==NULL){
			cout<< "NULL ";
			continue;
		}
		else {
			cout<<tmp->val<<" ";
		}
		
		count--;
		
		if ((tmp->left)==NULL)cnt_null++;
		if ((tmp->right)==NULL)cnt_null++;
		
				q.push(tmp->left);
		q.push(tmp->right);

		if (count==0)
		{
			if (cnt_null==2*tmp_cnt)break; // 如果当前节点NULL的个数是节点个数的2倍，那说明以及到达终点了。
			//cout<<"cnt_null: "<<cnt_null<<" "<<"tmp_cnt: "<<tmp_cnt<<endl;
			
				tmp_cnt = q.size()-cnt_null;
				count = q.size()-cnt_null;
				
				cnt_null = 0;
		}
		

		
		
	}
	
}