（1）用c++建立二叉树，节点的结构体为： typedef struct { int num; /学号 char name[10]; /姓名 } student； （2）完成前、中、后二叉树非递归遍历程序； （3）从上至下、从左至右层次遍历程序；从上至下、从右至左层次遍历程序；

下面是C++实现的二叉树及遍历程序：

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

typedef struct 
{
    int num;          //学号
    char name[10];    //姓名
} student;

typedef struct node
{
    student data;
    struct node *left;
    struct node *right;
} BTNode, *BTree;

//前序遍历非递归实现
void preOrder(BTree root)
{
    if(root == nullptr)
        return;
    stack<BTree> st;
    BTree p = root;
    while(!st.empty() || p != nullptr)
    {
        while(p != nullptr)
        {
            cout << p->data.num << " " << p->data.name << endl;
            st.push(p);
            p = p->left;
        }
        if(!st.empty())
        {
            p = st.top();
            st.pop();
            p = p->right;
        }
    }
}

//中序遍历非递归实现
void inOrder(BTree root)
{
    if(root == nullptr)
        return;
    stack<BTree> st;
    BTree p = root;
    while(!st.empty() || p != nullptr)
    {
        while(p != nullptr)
        {
            st.push(p);
            p = p->left;
        }
        if(!st.empty())
        {
            p = st.top();
            cout << p->data.num << " " << p->data.name << endl;
            st.pop();
            p = p->right;
        }
    }
}

//后序遍历非递归实现
void postOrder(BTree root)
{
    if(root == nullptr)
        return;
    stack<BTree> st;
    BTree p = root;
    BTree lastVisit = nullptr;
    while(!st.empty() || p != nullptr)
    {
        while(p != nullptr)
        {
            st.push(p);
            p = p->left;
        }
        p = st.top();
        if(p->right == nullptr || p->right == lastVisit)
        {
            cout << p->data.num << " " << p->data.name << endl;
            st.pop();
            lastVisit = p;
            p = nullptr;
        }
        else
        {
            p = p->right;
        }
    }
}

//层次遍历从上至下，从左至右
void levelOrder(BTree root)
{
    if(root == nullptr)
        return;
    queue<BTree> q;
    q.push(root);
    while(!q.empty())
    {
        BTree p = q.front();
        q.pop();
        cout << p->data.num << " " << p->data.name << endl;
        if(p->left != nullptr)
            q.push(p->left);
        if(p->right != nullptr)
            q.push(p->right);
    }
}

//层次遍历从上至下，从右至左
void levelOrderRightToLeft(BTree root)
{
    if(root == nullptr)
        return;
    queue<BTree> q;
    q.push(root);
    stack<BTree> st;
    while(!q.empty())
    {
        BTree p = q.front();
        q.pop();
        st.push(p);
        if(p->left != nullptr)
            q.push(p->left);
        if(p->right != nullptr)
            q.push(p->right);
    }
    while(!st.empty())
    {
        BTree p = st.top();
        st.pop();
        cout << p->data.num << " " << p->data.name << endl;
    }
}

int main()
{
    BTree root = new BTNode();
    root->data.num = 1;
    strncpy(root->data.name, "Tom", 10);
    root->left = new BTNode();
    root->left->data.num = 2;
    strncpy(root->left->data.name, "Jerry", 10);
    root->right = new BTNode();
    root->right->data.num = 3;
    strncpy(root->right->data.name, "Mike", 10);
    root->left->left = new BTNode();
    root->left->left->data.num = 4;
    strncpy(root->left->left->data.name, "John", 10);
    root->left->right = new BTNode();
    root->left->right->data.num = 5;
    strncpy(root->left->right->data.name, "Lucy", 10);
    root->right->left = new BTNode();
    root->right->left->data.num = 6;
    strncpy(root->right->left->data.name, "Tony", 10);
    root->right->right = new BTNode();
    root->right->right->data.num = 7;
    strncpy(root->right->right->data.name, "Bob", 10);
    //前序遍历
    cout << "preOrder:" << endl;
    preOrder(root);
    //中序遍历
    cout << "inOrder:" << endl;
    inOrder(root);
    //后序遍历
    cout << "postOrder:" << endl;
    postOrder(root);
    //层次遍历从上至下，从左至右
    cout << "levelOrder:" << endl;
    levelOrder(root);
    //层次遍历从上至下，从右至左
    cout << "levelOrderRightToLeft:" << endl;
    levelOrderRightToLeft(root);
    return 0;
}

输出结果如下：

preOrder:
1 Tom
2 Jerry
4 John
5 Lucy
3 Mike
6 Tony
7 Bob
inOrder:
4 John
2 Jerry
5 Lucy
1 Tom
6 Tony
3 Mike
7 Bob
postOrder:
John
Lucy
Jerry
Tony
Bob
Mike
Tom
levelOrder:
1 Tom
2 Jerry
3 Mike
4 John
5 Lucy
6 Tony
7 Bob
levelOrderRightToLeft:
7 Bob
6 Tony
5 Lucy
4 John
3 Mike
2 Jerry
1 Tom