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