Balanced binary tree (C language)

I think this is the most understandable version, adapted from the internet, the source forgot, I complete his code, and changed the style to my,, the function really does not like a brain of all throw above,. ，，，。

It is recommended to read these two articles:

Http://www.cnki.net/KCMS/detail/detail.aspx? Queryid=2&currec=1&recid=&filename=gwdt201034126&dbname=cjfdn0911&dbcode=cjfr&pr= CJFR2010; Cfjd2010;&urlid=&yx=&uid= Weevrecwsljhsldra1finlpyaknurfryd0tktukwazllww5tbjyzyxfvv3pdk3pwyjbqn3zznjdkqtlvqkpwn3l3pt0=$9a4hf_ yauvq5obgvaqnkpcycejkensw4iqmovwhtwkf4vypohbkxjw!! &v=mdayotjgeursvzd2sulqclblckc0sdliuhe0nuhzb1i4zvgxthv4wvm3rggxvdnxvhjxttfgcknvukw2zvplzhe=

Http://www.cnblogs.com/fornever/archive/2011/11/15/2249492.html

#include <stdio.h>#include<stdlib.h>#defineEH 0/* Equal height */#defineLH 1/* Left high */#defineRH-1/* Right high */typedefintElemtype;/*Data Type*/typedefstructbitree{elemtype data; /*Data Elements*/    intBF;/*Balance Factor*/    structBitree *lchild,*rchild;/*left and right child hands*/}*Bitree,bitreenode;intINSERTAVL (Bitree *t,elemtype E,BOOL*taller);voidLeftbalance (Bitree *T);voidRightbalance (Bitree *T);voidR_rotate (Bitree *T);voidL_rotate (Bitree *T);BOOL*taller= (BOOL*)malloc(sizeof(BOOL));intMainvoid){    intdata; Bitree T=NULL;  while(1) {printf ("Enter the number (zero to exit):"); scanf ("%d",&data); if(0==data) Break; Insertavl (&T,data,taller); }                return 0;}/*If there are no nodes with the same key code as E in the balanced two-fork sort tree T, insert a data element of E*//*new node, and reverse back to 1, or reverse back to 0. If the two-fork sorting tree loses its balance due to insertion, the balance rotation is processed,*/        /*Boolean variable taller reflects T-length height or not*/    intINSERTAVL (Bitree *t,elemtype E,BOOL*taller) {    if(!*t)/*Insert new node, tree "long height", set taller to Ture*/    {        (*t) = (bitree)malloc(sizeof(Bitreenode)); (*t)->data =e; (*t)->lchild = (*t)->rchild =NULL; (*t)-&GT;BF =EH; *taller =true; }    Else    {        if(e== (*t)->data)/*There are nodes in the tree that have the same key code as E, without inserting*/        {            *taller =false; return 0; }            if(e< (*t),data) {            if(! Insertavl (& (*t)->lchild,e,taller))return 0;/*not inserted*/            if(*taller)Switch((*t)BF) {                     CaseEh:/*originally left and right sub-tree, such as high, due to the increase of the left subtree to increase the tree*/                    (*T)->bf=LH; *taller=true;  Break;  CaseLh:/*the left sub-tree is tall and needs to be left balanced.*/leftbalance (T); *taller=false;  Break;  CaseRh:/*originally right sub-tree high, so that the left and right sub-tree, such as high*/                    (*T)->bf=EH; *taller=false;  Break; }                    }        Else        {            if(! Insertavl (& (*t)->rchild,e,taller))return 0;/*not inserted*/            if(*taller)Switch((*t)BF) {                     CaseEh:/*originally left and right sub-tree, such as high, because the right subtree increased to increase the tree*/                    (*T)->bf=RH; *taller=true;  Break;  CaseLh:/*originally left sub-tree high, so that the left and right sub-tree, such as high*/                    (*T)->bf=EH; *taller=false;  Break;  CaseRh:/*the right sub-tree is tall and needs to be treated with right balance.*/rightbalance (T); *taller=false;  Break; }        }    }    return 1;}/*The *p point is the root of the sub-tree, the left balance rotation processing, after processing, the *p points to the new root of the subtree*/voidLeftbalance (Bitree *T) {Bitree L= (*t)->lchild,lr;/*L Point to *t Zogen Junction.*/    Switch(L-&GT;BF)/*check L balance and handle accordingly*/    {         CaseLh:/*The new node is inserted in the left subtree of the *p Zuozi, and it needs to be rotated by single right .*/            (*T)->bf=l->bf=EH;             R_rotate (T);  Break;  CaseEh:/*originally left and right sub-tree, such as high, due to the increase of the left subtree to increase the tree*/            (*T)->bf=lh;//eh seems to have no need to write here.*taller=true;  Break;  CaseRh:/*The new node is inserted in the right sub-tree of the left child of *t, and must be left and right double-spin .*/Lr=l->rchild;/*Lr points to *p left child's right subtree node .*/                Switch(LR-&GT;BF)/*correction of the balance factor of *t and its left sub-tree*/            {                 CaseLH: (*t)-&GT;BF =RH; L-&GT;BF =EH;  Break;  CaseEH: (*t)-&GT;BF = l->bf=EH;  Break;  CaseRH: (*t)-&GT;BF =EH; L-&GT;BF =LH;  Break; } Lr-&GT;BF =EH; L_rotate (&AMP;L);/*left rotation of the *t Zuozi*/r_rotate (T); /*right rotation treatment of *t*/    }}//here and leftbalance a truth, try to write your ownvoidRightbalance (Bitree *T) {Bitree Lr= (*t)rchild,l; Switch(lr->BF) {         CaseEH:*taller =true; (*t)-&GT;BF =RH;  Break;  CaseRH: (*T)->bf=lr->bf=EH;            L_rotate (T);  Break;  Caselh:l= lr->Lchild; Switch(l->BF) {                 CaseEH: (*T)->bf=lr->bf=EH;  Break;  CaseRH:LR->bf=EH; (*t)-&GT;BF =LH;  Break;  CaseLH: (*t)-&GT;BF =LH; Lr-&GT;BF =EH;  Break; } L-&GT;BF =EH; R_rotate (&Lr);                L_rotate (T); }}/*The sub-tree, which is the root of the node pointed by *t, is processed by the right single rotation, and after processing, the node *t points to is the new root of the subtree.*/voidR_rotate (Bitree *T) {Bitree L= (*t)->lchild;/*L Point to *t Zogen Junction.*/    (*T)->lchild=l->rchild;/*L's right son rime the left subtree of *t.*/L->rchild=*t; *t=l;/**l point to the new root node*/}/*The *p point is the root of the sub-tree, the left single rotation processing, after processing, the *p points to the new root of the subtree*/voidL_rotate (Bitree *T) {Bitree Lr= (*t)->rchild;/*Lr points to the *t right subtree node.*/    (*T)->rchild=lr->lchild;/*L's left child rime to *p right subtree*/Lr->lchild=*T; *T=LR;/**l point to the new root node*/}

Balanced binary tree (C language)