谁能帮我把这个程序注释一下，越详细越好，谢谢了，我重重有赏

Question

#include<stdio.h>
#include<stdlib.h>
#define MAXSIZE 100
typedef char datatype ;
typedef struct node
{
datatype data;
struct node *lchild,*rchild;
}treenode;

typedef treenode *BiTree;

typedef struct stack
{
BiTree data[MAXSIZE];
int tag[MAXSIZE];
int top;
}seqstack;

BiTree pop(seqstack *s)
{
if(s->top!=0)
{
s->top--;
return s->data[s->top];
}
else return NULL;
}
void push(BiTree t,seqstack *s)
{
s->data[s->top]=t;s->top++;
}

void creatree(BiTree &T)
{
char ch;
ch=getchar();
if(ch=='#') T=NULL;
else
{
T=(BiTree)malloc(sizeof(treenode));
T->data=ch;
creatree(T->lchild);
creatree(T->rchild);
}
}

void preorder(BiTree T)
{
if(T)
{
printf("%c",T->data);
preorder(T->lchild);
preorder(T->rchild);
}
}
void inorder(BiTree T)
{
if(T)
{
inorder(T->lchild);
printf("%c",T->data);
inorder(T->rchild);
}
}

int find(BiTree T,BiTree x,seqstack *s)
{

s->top=0;
while(T||s->top)
{

if(T)
{
s->tag[s->top]=0;
push(T,s);
T=T->lchild;
}
else
{
if(s->top&&s->tag[s->top-1])
{
T=pop(s);
if(T&&T->data==x->data)
return s->top;
T=NULL;

}
else
{
T=s->data[s->top-1];
s->tag[s->top-1]=1;
T=T->rchild;
}

}

}

return -1;
}

void grandpar(seqstack *sq,seqstack *sp)
{
while(sq->top!=sp->top)
{
if((sq->top-1)>(sp->top-1))
{
pop(sq);
}
else
{
pop(sp);
}
}
while(sp->top)
{
if(sq->data[sq->top-1]==sp->data[sp->top-1])
{
printf("%4c",sq->data[sq->top-1]->data);
}
pop(sq);
pop(sp);
}
}

main()
{
BiTree root,p,q;
seqstack *sp,*sq;
int x,y;
root=(BiTree)malloc(sizeof(treenode));
p=(BiTree)malloc(sizeof(treenode));
q=(BiTree)malloc(sizeof(treenode));
sq=(seqstack *)malloc(sizeof(seqstack));
sp=(seqstack *)malloc(sizeof(seqstack));
printf("请用前序遍历创建一棵二叉树：");
creatree(root);
printf("该二叉树为");
preorder(root);
printf("\n请输入要查找共同祖先的两个结点：");
getchar();

scanf("%c%c",&p->data,&q->data);

x=find(root,p,sp);
y=find(root,q,sq);
printf("这两个结点的共同祖先为");
grandpar(sq,sp);
}

Answer 1

#include<stdio.h>
#include<stdlib.h>
#define MAXSIZE 100

// 树结构定义
typedef char datatype ;
typedef struct node {
datatype data;
struct node *lchild, *rchild;
} treenode;

typedef treenode *BiTree;

// 栈
typedef struct stack {
BiTree data[MAXSIZE];
int tag[MAXSIZE];
int top;
} seqstack;

// 出栈操作
BiTree pop (seqstack *s)
{
// 若栈顶不为空
if (s->top != 0) {
// 减一个
s->top--;
// 返回减掉的data
return s->data[s->top];
} else return NULL;
// 否则返回空
}
// 入栈操作
void push (BiTree t, seqstack *s)
{
// 把t放上栈顶
s->data[s->top] = t;
// 栈顶递增
s->top++;
}

// 生成树
void creatree (BiTree &T)
{
char ch;
// 读取字符
ch = getchar();
// 输入#结束输入
if (ch == '#') T = NULL;
// 录入
else {
// 分配
T = (BiTree) malloc (sizeof (treenode));
// 放入data
T->data = ch;
// 继续二叉
creatree (T->lchild);
creatree (T->rchild);
}
}

// 前序遍历
void preorder (BiTree T)
{
if (T) {
// 前序、中序的区别自个弄清概念
printf ("%c", T->data);
preorder (T->lchild);
preorder (T->rchild);
}
}
// 中序遍历
void inorder (BiTree T)
{
if (T) {
// 顺藤摸瓜
inorder (T->lchild);
printf ("%c", T->data);
inorder (T->rchild);
}
}

// 查找树
int find (BiTree T, BiTree x, seqstack *s)
{
// 栈顶复0
s->top = 0;
// 若T不为空
while (T || s->top) {
// 能进来当然不空
if (T) {
s->tag[s->top] = 0;
push (T, s);
T = T->lchild;
// 这函数似乎乱七八糟的，这程序是谁写的，你读别人的吗，这是个正常的程序吗
} else {
if (s->top && s->tag[s->top-1]) {
T = pop (s);
if (T && T->data == x->data)
return s->top;
T = NULL;

} else {
T = s->data[s->top-1];
s->tag[s->top-1] = 1;
T = T->rchild;
}

}

}

return -1;
}

// 查找共同祖先
void grandpar (seqstack *sq, seqstack *sp)
{
// 先把栈顶砍平
while (sq->top != sp->top) {
if ( (sq->top - 1) > (sp->top - 1)) {
pop (sq);
} else {
pop (sp);
}
}
// 平了开始比较对应元素
while (sp->top) {
if (sq->data[sq->top-1] == sp->data[sp->top-1]) {
// 若找到相同的就打印
printf ("%4c", sq->data[sq->top-1]->data);
}
pop (sq);
pop (sp);
}
}

int main()
{
BiTree root, p, q;
seqstack *sp, *sq;
int x, y;
root = (BiTree) malloc (sizeof (treenode));
p = (BiTree) malloc (sizeof (treenode));
q = (BiTree) malloc (sizeof (treenode));
sq = (seqstack *) malloc (sizeof (seqstack));
sp = (seqstack *) malloc (sizeof (seqstack));
printf ("请用前序遍历创建一棵二叉树：");
creatree (root);
printf ("该二叉树为");
preorder (root);
printf ("\n请输入要查找共同祖先的两个结点：");
getchar();

scanf ("%c%c", &p->data, &q->data);

x = find (root, p, sp);
y = find (root, q, sq);
printf ("这两个结点的共同祖先为");
grandpar (sq, sp);
}

追问

废话，当然不是我写的啦，要是我写的怎么可能不会注释，不过看你给我回答了，就当做采纳了