请实现一个函数按照之字形打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右至左的顺序打印，第三行按照从左到右的顺序打印，其他行以此类推。

// test20.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include<iostream>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<cstring>
#include<string.h>
#include<deque>

using namespace std;



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) {
		vector<vector<int>> vec;
		vector<int> v;
		deque<TreeNode *>  parent;//用来打印父节点
		deque<TreeNode *>  child;//用来存储子节点
		int flag = 0;//用来标注是奇数行还是偶数行；奇数行从左像右打印；偶数行从右向左打印；

		if (pRoot == NULL) return{};
		parent.push_back(pRoot);
		while (!parent.empty ())
		{
			++flag;
			for (auto it = parent.begin();it != parent.end();it++)
			{
				v.push_back((*it)->val);
			}
			vec.push_back(v);
			v.clear();
			if (flag % 2==1)
			{
				while (!parent.empty())
				{
					if (parent.back()->right != NULL)	child.push_back(parent.back()->right);
					if(parent.back()->left!=NULL)	child.push_back(parent.back()->left);
					parent.pop_back();
				}
			}
			else
			{
				while (!parent.empty())
				{
					if (parent.back()->left != NULL)	child.push_back(parent.back()->left);
					if (parent.back()->right != NULL)	child.push_back(parent.back()->right);
					parent.pop_back();
				}
			}
			while (!child.empty ())
			{
				parent.push_back(child.front());
				child.pop_front();
			}
			
		}
		return vec;
	}
	int  NodeCount(TreeNode *T)
	{
		if (T == NULL) return 0;
		else
		{
			return NodeCount(T->left) + NodeCount(T->right) + 1;
		}
	}

	int count_0=0, count_1=0, count_2=0;
	void NodeCoutNUM(TreeNode *T)
	{
		if (T == NULL) return;
		if (T->left == NULL&&T->right == NULL)
		{
			++count_0;
		}
		if (T->left != NULL&&T->right == NULL)
		{
			++count_1;
			NodeCoutNUM(T->left);
	
		}
		if (T->left == NULL&&T->right != NULL)
		{
			++count_1;
			NodeCoutNUM(T->right);
		}

		if (T->left != NULL&&T->right != NULL)
		{
			++count_2;
			NodeCoutNUM(T->left);
			NodeCoutNUM(T->right);
		}
	
	}

	

	void createBiTree(TreeNode* &T)
	{
		int num;
		cin >> num;
		if (num==0) return;
		else
		{
			T = new TreeNode(num);
			createBiTree(T->left);
			createBiTree(T->right);
		}
	}
	void preOrderTraver(TreeNode *T)
	{
		if (T == NULL) return;
		else
		{
			cout << T->val << "  ";
			preOrderTraver(T->left);
			preOrderTraver(T->right);
		}
	}

};
int main()
{
	
	Solution so;
	TreeNode *T=NULL;

	so.createBiTree(T);
	cout << "创建T成功！" << endl;
	cout << "前序遍历二叉树的结果是：" << endl;
	so.preOrderTraver(T);
	cout << endl;
	
	so.NodeCoutNUM(T);
	cout << "总的节点个数：" << so.NodeCount(T) << endl;
	cout << "度为 0 的节点个数：" <<so.count_0 <<endl;
	cout << "度为 1 的节点个数：" << so.count_1 << endl;
	cout << "度为 2 的节点个数：" << so.count_2 << endl;

	vector<vector<int> > vec =so.Print(T);
	for (auto it = vec.begin();it != vec.end();++it)
	{
		for (auto i = it->begin();i != it->end();++i)
		{
			cout << *i << "  " ;
		}
		cout << endl;
	}
	
	cout << endl;
	return 0;
}

注意：使用到容器 双向队列；
另外可以使用 reserve，但是这种方法效率比较低；
此方法中还有计算二叉树的节点个数，度数为0的节点个数，度数为1的节点个数，度数为2的节点个数的算法