二叉树主函数和子函数输出问题求助！！
各位大神看看为什么main函数里面out(T)和run函数的out(T)输出的结果会不一样#include"stdio.h"
#include"stdlib.h"
#include"math.h"
#include"conio.h"
struct matrix{
int l; //记录矩阵长度
char title[22][22]; //记录每位的字母
int f[22]; //记录每列字母个数
}; //记录矩阵状态
struct tree{
char *names;
float num;
struct tree *left;
struct tree *right;
}; //记录二叉树
float usk[4][22][22]={
//A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V
5,0,3,0,0,0,0,5,0,0,0,3,1,0,0,0,5,0,0,0,0,0,//s_A
3,5,0,0,5,0,1,0,0,3,0,0,0,0,0,0,0,0,0,0,0,0,//B
0,0,5,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,0,0,//C
0,5,0,5,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,//D
0,0,0,0,5,0,0,0,0,0,0,3,0,0,0,0,0,0,3,0,0,0,//E
1,0,0,3,0,5,0,0,0,0,0,0,0,0,5,0,0,0,0,0,0,0,//F
0,0,0,0,0,0,5,0,0,5,0,0,0,0,0,0,0,0,0,0,0,0,//G
0,0,3,0,0,0,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,//H
0,0,0,0,0,0,0,0,5,0,0,0,0,3,0,0,1,0,0,0,0,1,//I
1,0,0,0,3,0,0,0,0,5,0,0,0,0,0,0,0,0,0,0,0,0,//J
0,0,0,0,0,0,0,0,0,0,5,0,0,0,0,0,0,0,0,0,0,0,//K
0,0,5,0,0,0,3,0,0,0,0,5,0,0,0,0,0,3,0,0,0,0,//L
0,0,0,0,0,0,0,0,0,0,0,0,5,0,0,0,0,0,0,3,0,1,//M
0,0,0,0,5,0,0,0,3,3,0,0,0,5,0,5,0,0,0,0,0,0,//N
0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,0,0,0,0,0,0,0,//O
0,3,0,0,0,0,0,0,0,0,0,5,0,0,0,5,0,0,0,0,0,0,//P
0,0,3,0,0,0,0,0,0,0,0,0,0,0,0,0,5,0,0,0,0,0,//Q
0,0,0,1,0,3,0,0,5,0,0,3,0,0,0,0,0,5,0,0,0,0,//R
0,0,0,0,0,0,0,0,0,0,0,0,0,0,3,0,0,0,5,0,5,0,//S
1,0,3,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,5,0,0,//T
0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,5,5,//U
0,0,0,0,0,0,0,0,0,0,0,0,0,5,0,0,0,3,0,0,0,5 //V
};//初始矩阵
void out(tree ls);
matrix ma[2];
tree ercha[22];
tree *t[22]; //用于指向所有字母
//用于每次更新矩阵所排出的关联度
//初始化矩阵和树
void install(){
int i,j;
ma[0].l=ma[1].l=22;
for(i=0;i<22;i++){
ma[1].f[i]=ma[0].f[i]=1;
ma[1].title[i][0]=ma[0].title[i][0]=65+i;
}//初始化每列字母信息
for(i=0;i<22;i++){
ercha[i].names=&ma[1].title[i][0];
ercha[i].num=100;
ercha[i].left=ercha[i].right=NULL;
t[i]=&ercha[i];
}//为每个字母构造一棵树
for(i=0;i<22;i++){
for(j=0;j<22;j++)
usk[3][i][j]=usk[0][i][j];
}//保存初始矩阵
}
//计算初始关联度
void count(){
int i,j,k;
float m,n; //分子，分母
for(i=0;i<22;i++)
for(j=i+1;j<22;j++){
m=0;
n=0;
for(k=0;k<22;k++){
n+=usk[3][k][i]+usk[3][k][j];
if(usk[3][k][i] && usk[3][k][j])
m+=usk[3][k][i]+usk[3][k][j];
}
usk[2][j][i]=usk[1][j][i]=m/n; //将初始关联度保存到第三个矩阵
}
}
//更新矩阵
void run(int s,tree *T){
int i,j,a,b,c;
int k=s%2;
int m=1,n=0;
float unite[22]; //保存联合关联度
for(i=0;i<22;i++){
for(j=0;j<22;j++)
usk[(s+1)%2][i][j]=0;
} //清空另一个矩阵
for(i=0;i<MA[K].L;I++){
for(j=i+1;j<MA[K].L;J++){
if(usk[k][j][i]>usk[k][m][n]){
m=j;
n=i; //记住最大的关联度的行与列
}
}
}
T->num=usk[k][m][n]; //保留最大关联度
if(m<N){
i=m;
m=n;
n=i;
}
T->left=t[m];
T->right=t[n];
out(*T);
system("pause");
//排序关联度最大的两列
printf("新的联合关联度为：");
for(i=0,j=0;i<MA[K].L;I++){
if(i!=m && i!=n){
unite[j]=(usk[k][i][m]+usk[k][i][n]+usk[k][m][i]+usk[k][n][i])/2;
printf(" %.3f",unite[j]);
j++;
}
}
printf("\n");
//去掉排序关联度最大的两列，并把后面的往前推进
for(i=0;i<MA[K].L;I++){
if(i==m || i==n)
continue;
if(i>m)
c=2;
else if(i>n)
c=1;
else c=0;
t[i-c]=t[i];
for(j=0;j<MA[K].L;J++){
if(j==m || j==n)
continue;
if(i>m)
a=2;
else if(i>n)
a=1;
else a=0;
if(j>m)
b=2;
else if(j>n)
b=1;
else b=0;
usk[(s+1)%2][i-a][j-b]=usk[k][i][j];//将剩下的关联度保存到另一个矩阵里（压缩矩阵）
}
}
t[i-2]=T;//将取出的行和列的字母转到最后的行和列
for(i=0;i<MA[K].L;I++)
usk[(s+1)%2][ma[k].l-2][i]=unite[i]; //将联合关联度保存到矩阵中
for(i=0;i<MA[K].L;I++){
if(i==m || i==n)
continue;
if(i>m)
a=2;
else if(i>n)
a=1;
else a=0;
for(j=0;j<MA[K].F[I];J++)
ma[(s+1)%2].title[i-a][j]=ma[k].title[i][j];
ma[(s+1)%2].f[i-a]=ma[k].f[i];
}
b=0;
for(j=0;j<MA[K].F[M];J++){
ma[(s+1)%2].title[i-2][b]=ma[k].title[m][j];
b++;
}
for(j=0;j<MA[K].F[N];J++){
ma[(s+1)%2].title[i-2][b]=ma[k].title[n][j];
b++;
}
ma[(s+1)%2].f[i-2]=ma[k].f[m]+ma[k].f[n];
ma[(s+1)%2].l=ma[k].l-1;
}
void display(int s){
int i,j;
printf("\nThe Maxtrix %d\n",ma[s%2].l);
for(i=0;i<MA[S%2].L;I++){
for(j=0;j<MA[S%2].F[I];J++)
printf("%c",ma[s%2].title[i][j]);
putchar('\n');
}
getchar();
}
void out(tree s){
tree ls=s;
if(ls.left && ls.right){
out(*ls.left);
out(*ls.right);}
else if(ls.num==100)
printf(" %s",ls.names);
}
void main(){
int i;
tree T;
getchar();
install();
count();
for(i=0;ma[i%2].l>2;i++){
run(i+1,&T);
out(T);
system("pause");
display(i);
}
out(T);
getchar();
}
