数据结构实验之查找二：平衡二叉树

来源：互联网发布：视频数码修复软件编辑：程序博客网时间：2024/06/16 00:11

大部分摘自，略有不同http://www.cnblogs.com/You0/p/4459719.html
Problem Description

根据给定的输入序列建立一棵平衡二叉树，求出建立的平衡二叉树的树根。
Input

输入一组测试数据。数据的第1行给出一个正整数N(n <= 20)，N表示输入序列的元素个数；第2行给出N个正整数，按数据给定顺序建立平衡二叉树。
Output

输出平衡二叉树的树根。
Example Input

588 70 61 96 120

Example Output

Hint

Author
xam

#include <iostream>#include <stdio.h>#include <cstring>#include <cstdlib>using namespace std;bool taller;enum wek{LH, EH, RH};struct  Node{    int data;    wek TQ;    Node *left;    Node *right;    Node():data(0),TQ(EH),left(NULL), right(NULL){}};Node *newNode(){return new Node;}int creat(Node *&T, int i);void leftbalance(Node *&T);void rightbalance(Node *&T);void l_xuan(Node *&T);void r_xuan(Node *&T);int main(){    int t;    while(~scanf("%d", &t))    {        Node *T=NULL;        while(t--)        {            int i;            scanf("%d", &i);            creat(T,i);        }        printf("%d\n", T->data);    }    return 0;}int creat(Node *&T, int i){    if(!T)    {        T=newNode();        taller=true;        T->data=i;    }    else    {        if(T->data==i)return 0;        else if(T->data>i)        {            if(!creat(T->left, i))return 0;            if(taller)            switch (T->TQ)            {                case LH:                    leftbalance(T);                    taller=false;                    break;                case EH:                    taller=true;                    T->TQ=LH;                    break;                case RH:                    T->TQ=EH;                    taller=false;                    break;            }        }        else        {            if(!creat(T->right, i))return 0;            if(taller)            switch (T->TQ)            {                case LH:                    T->TQ=EH;                    taller=false;                    break;                case EH:                    T->TQ=RH;                    taller=true;                    break;                case RH:                    rightbalance(T);                    taller=false;                    break;            }        }    }    return 1;}void leftbalance(Node *&T){    Node *&L=(T->left);    switch (L->TQ)    {        case LH:            T->TQ=EH;            L->TQ=EH;            r_xuan(T);            break;        case EH:            T->TQ=LH;            taller=true;            break;        case RH:            Node *&Lr=(L->right);            switch (Lr->TQ)            {                case RH:                    T->TQ=EH;                    L->TQ=LH;                    break;                case EH:                    T->TQ=EH;                    L->TQ=EH;                    break;                case LH:                    T->TQ=RH;                    L->TQ=EH;            }            Lr->TQ=EH;            l_xuan(L);            r_xuan(T);    }}void rightbalance(Node *&T){    Node *&R=T->right;    switch (R->TQ)    {        case EH:            T->TQ=RH;            taller=true;            break;        case RH:            T->TQ=EH;            R->TQ=EH;            l_xuan(T);            break;        case LH:            Node *&Rl=R->left;            switch (Rl->TQ)            {                case LH:                    T->TQ=EH;//这部分有小的改动，可能有错                    R->TQ=RH;                    break;                case EH:                    T->TQ=EH;                    R->TQ=EH;                    break;                case RH:                    R->TQ=EH;                    T->TQ=LH;                    break;            }            Rl->TQ=EH;            r_xuan(R);            l_xuan(T);            break;    }}void r_xuan(Node *&T){    Node *L=T->left;    T->left=L->right;    L->right=T;    T=L;}void l_xuan(Node *&T){    Node *R=T->right;    T->right=R->left;    R->left=T;    T=R;}

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