14_二叉树

#include<iostream>#include <string>#include <stack>using namespace std;struct MyStruct{    int Nodedata;    MyStruct *pLeft;    MyStruct *pRight;}BTree,*pBTree;//中序，前序，后序，递归遍历，非递归遍历（栈）//查找，修改，插入，排序,删除(太难)/*递归：在输出doc界面上，输出二叉树的形状*/void  show(MyStruct *proot  ,int n){    if (proot==nullptr)    {        return;    }     else    {        show(proot->pRight, n + 1);        for (int i = 0; i < n;i++)        {            cout << "   ";        }        cout << proot->Nodedata << endl;        show(proot->pLeft, n + 1);    }}/*递归：实现二叉树中序遍历*/void zhong(MyStruct *proot){    if (proot!=nullptr)    {        if (proot->pLeft!=nullptr)        {            zhong(proot->pLeft);        }        cout << " " << proot->Nodedata ;        if (proot->pRight!= nullptr)        {            zhong(proot->pRight);        }    }}/*递归：实现二叉树先序遍历*/void xian(MyStruct *proot){    if (proot!=nullptr)    {        cout << " " << proot->Nodedata ;        if (proot->pLeft!=nullptr)        {            xian(proot->pLeft);        }        if (proot->pRight!= nullptr)        {            xian(proot->pRight);        }    }}/*递归：实现二叉树后序遍历*/void hou(MyStruct *proot){    if (proot!=nullptr)    {        if (proot->pLeft!=nullptr)        {            hou(proot->pLeft);        }        if (proot->pRight!= nullptr)        {            hou(proot->pRight);        }        cout << " " << proot->Nodedata ;    }}/*利用数组实现二叉树的中序遍历*/void  stackzhong(MyStruct *proot){    MyStruct * pcurr = proot;//记录根节点    MyStruct * mystack[100];//指针数据    int top = 0;     while ( top!=0 || pcurr !=nullptr)     {         while (pcurr != nullptr)         {             mystack[top++] = pcurr;             pcurr = pcurr->pLeft;         }         if (top>0)         {             top--;             pcurr = mystack[top];             cout << "  " << pcurr->Nodedata << endl;             pcurr = pcurr->pRight;         }     }}/*利用栈实现二叉树的中序遍历*/void  stackzhongA(MyStruct *proot){    MyStruct * pcurr = proot;//记录根节点    stack<MyStruct *> mystack;    while (!mystack.empty() || pcurr != nullptr)    {        while (pcurr != nullptr)        {            mystack.push(pcurr);            pcurr = pcurr->pLeft;        }        if (!mystack.empty())        {            pcurr = mystack.top();            cout << "  " << pcurr->Nodedata << endl;            mystack.pop();            pcurr = pcurr->pRight;        }    }}/*递归：获取叶子节点数*/int getyenum(MyStruct *proot){    int left = 0;    int right = 0;    if (proot==nullptr)    {        return 0;    }    if (proot->pLeft==nullptr && proot->pRight==nullptr)    {        return 1;    }    left = getyenum(proot->pLeft);    right = getyenum(proot->pRight);    return left + right;}/*递归：求二叉树的深度*/int  getheight(MyStruct *proot){    int height = 0;    int left = 0;    int right = 0;    if (proot == nullptr)    {        return 0;    }    left = getheight(proot->pLeft);    right = getheight(proot->pRight);    height = left > right ? left : right;    return height + 1;}/*队列*/void ceng(MyStruct *proot){    if (proot ==nullptr)    {        return;    }    MyStruct * myq[100];    int tou = 0;    int wei = 0;    MyStruct * pcurr = nullptr;    myq[wei++] = proot;//存入队列第一个节点，入队    while (tou !=wei)    {        pcurr = myq[tou];        tou++;//出队        cout << pcurr->Nodedata << endl;        if (pcurr->pLeft!=nullptr)        {            myq[wei++] = pcurr->pLeft;//入队        }        if (pcurr->pRight != nullptr)        {            myq[wei++] = pcurr->pRight;//入队        }    }}/*递归：输入某个数，获取到他的父节点*/int getba(MyStruct *pRoot,int num){    if (pRoot==nullptr)    {        return 0;    }    if (pRoot->pLeft!=nullptr && pRoot->pLeft->Nodedata==num)    {        return pRoot->Nodedata;    }    if (pRoot->pRight != nullptr && pRoot->pRight->Nodedata == num)    {        return pRoot->Nodedata;    }    getba(pRoot->pLeft, num);    getba(pRoot->pRight, num);}/*递归：得到它的左兄弟节点*/int  getleft(MyStruct *pRoot, int num){    if (pRoot == nullptr)    {        return 0;    }    if (pRoot->pRight && pRoot->pRight->Nodedata == num)    {        if (pRoot->pLeft)        {            return  pRoot->pLeft->Nodedata;        }    }    getleft(pRoot->pLeft, num);    getleft(pRoot->pRight, num);}/*利用数组模拟栈，查找二叉树中的最大值*/int findmax(MyStruct *proot){    int max = -99999;    MyStruct * pcurr = proot;//记录根节点    MyStruct * mystack[100];//指针数据    int top = 0;    while (top != 0 || pcurr != nullptr)    {        while (pcurr != nullptr)        {            mystack[top++] = pcurr;            pcurr = pcurr->pLeft;        }        if (top > 0)        {            top--;            pcurr = mystack[top];            if (max< pcurr->Nodedata)            {                max = pcurr->Nodedata;            }            pcurr = pcurr->pRight;        }    }    return max;}/*递归：插入数据*/MyStruct * insertnode(MyStruct *proot,int num){    if (proot==nullptr)    {        MyStruct *pnew = new MyStruct;        pnew->Nodedata = num;        pnew->pLeft = nullptr ;        pnew->pRight = nullptr ;        proot = pnew;    }     else  if ( num <=  proot->Nodedata)    {        proot->pLeft = insertnode(proot->pLeft, num);    }    else     {        proot->pRight = insertnode(proot->pRight, num);    }    return proot;}void main01(){    MyStruct *pRoot;//根    MyStruct s1 = { 0 , nullptr , nullptr} ;    MyStruct s2 = { 0 , nullptr , nullptr} ;    MyStruct s3 = { 0 , nullptr , nullptr} ;    MyStruct s4 = { 0 , nullptr , nullptr} ;    MyStruct s5 = { 0 , nullptr , nullptr} ;    MyStruct s6 = { 0 , nullptr , nullptr} ;    MyStruct s7 = { 0 , nullptr , nullptr} ;    MyStruct s8 = { 0 , nullptr , nullptr} ;    MyStruct s9 = { 0 , nullptr , nullptr} ;    pRoot = &s1;    s1.Nodedata = 1;    s2.Nodedata = 2;    s3.Nodedata = 3;    s4.Nodedata = 4;    s5.Nodedata = 5;    s6.Nodedata = 6;    s7.Nodedata = 7;    s8.Nodedata = 8;    s9.Nodedata = 9;    s1.pLeft = &s2;    s1.pRight = &s3;    s2.pLeft = &s4;    s2.pRight = &s5;    s3.pLeft = &s6;    s3.pRight = &s7;    s4.pLeft = &s8 ;    s4.pRight = nullptr ;    s8.pRight = &s9 ;    s8.pLeft = nullptr ;    show( pRoot , 1) ;    cout<<endl ;    zhong(pRoot) ;    cout<<endl ;    xian(pRoot) ;    cout<<endl ;    hou(pRoot) ;    cout<< endl << getyenum( pRoot );    cout<< endl << getheight( pRoot ) ;     ceng( pRoot ) ;    cin.get();}void mainA(){    MyStruct *pRoot;//根    MyStruct sarray[100];     /*数组的每个元素都是二叉树上的一个节点*/    pRoot = sarray;    for (int i = 1; i <= 100;i++)    {        sarray[i].Nodedata = i;    }    /*给数组的每个元素的前驱和后继赋值，    也就是让数组形成逻辑上的二叉树。    究竟有多少个节点需要赋值呢？    实际上只有最下面的一层不需要赋值，    倒数第二层需要赋值（有些不要）    所以在循环的内部还要加限定条件。    所以这里的大循环应该是倒数第二层到    第一层之间的节点总个数（设为x），    则最后一层的节点个数为（x+1），    条件为：x+x+1<=100。这里取相近的50.    反正在循环的内部还会限制条件*/    for (int i = 0; i <= 50;i++)    {        /*正如上面所说的，由于设置的总节点个数        不一定能设整个二叉树成为满二叉树，所以        在倒数第二层这里最右边有些节点是不会挂上        前驱和后继的。        还有可能只是挂上了前驱没后继，        所以需要加限制条件来赋值。        条件要求：假设要挂前驱和后继的节点为x，        则其挂载的前驱为：2x+1（这里根据i从0开始推倒出来的）；        后继为：2x+2；        可得：        只要满足2x+1<=99 即可挂上前驱;        只要满足2x+2<=99 即可挂上后继*/        if (i<=(99-1)/2)        {            sarray[i].pLeft = &sarray[2 * i + 1];        }        if (i<=(99-2)/2)        {            sarray[i].pRight = &sarray[2 * i + 2];        }    }    show(pRoot, 1);    cin.get();}void main(){    MyStruct *pRoot=nullptr;//根    for (int i = 0; i < 10; i+=2)    {        pRoot = insertnode(pRoot, i);    }    for (int i = 5; i >=0; i-=2)    {        pRoot = insertnode(pRoot, i);    }    show(pRoot, 1);    cin.get();}