学校OJ显示RUNTIME ERROR，VS2017调试无误

代码我放在下面了
恳请各位大佬帮忙看看（能上优快云的除我都是大佬哈哈哈）
连续写三个题目都是RUNTIME ERROR！！！
找指针越界找的心力交瘁也没找出来！！！
呜呜呜
这是第一题的题目

计算二叉树的高度和结点数
Description:
二叉树是非常重要的树形数据结构，根据该树的先序、中序或后序遍历序列可以建立一棵二叉树。例如输入先序遍历序列A B # D # # C E # # F # #可以建立图1019-1所示的二叉树，这里用#代表空树或空子树（另一种说法：若无孩子结点，则用#代替），如图1019-2。

图1019-1

图1019-2

请实现基于遍历的二叉树运算：求高度、计算结点数目 Input: 二叉树的先序遍历序列，用#代表空树或空子树。

Output: 共五行

前三行依次输出先序、中序和后序遍历序列，

第四行输出二叉树的高度，

第五行依次输出二叉树总结点数目、叶子结点数目、度为1的结点数目。

Sample Input:
A B # D # # C E # # F # #
Sample Output:
PreOrder:A B D C E F
InOrder: B D A E C F
PostOrder: D B E F C A
3
6 3 1

下面是我写的代码

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define ERROR 0
#define OK 1

static i = 0;                                               //遍历数组所需变量

typedef struct BinaryTreeNode
{
	char Data;
	struct BinaryTreeNode *lChild, *rChild;
}BinaryTreeNode;
typedef struct BinaryTree
{
	BinaryTreeNode *root;
}BT;
void Input(char *a);
BinaryTreeNode* PreCreateBinaryTree(BinaryTreeNode *t, char *a);
void PreOrderTransverse(BinaryTreeNode *t);
void InOrderTransverse(BinaryTreeNode *t);
void PostOrderTransverse(BinaryTreeNode *t);
void getLeavesCount(BinaryTreeNode* T, int *count);
int Height(BinaryTreeNode *t)             //强
{
	int height;
	int leftHeight = 0;
	int rightHeight = 0;
	if (t->lChild != NULL)
		leftHeight = Height(t->lChild);//得到左子树高度
	if (t->rChild != NULL)
		rightHeight = Height(t->rChild);//得到右子树高度
	height = (leftHeight > rightHeight ? leftHeight : rightHeight) + 1;//返回高度
	return height;
}
void OneChildCount(BinaryTreeNode* T, int *count);
int main()
{
	BT t;
	int count = 0, index;
	char a[50];
	Input(a);
	t.root = NULL;
	t.root = PreCreateBinaryTree(&t.root, a);
	printf("PreOrder:");
	PreOrderTransverse(t.root);
	printf("\n");
	printf("InOrder:");
	InOrderTransverse(t.root);
	printf("\nPostOrder:");
	PostOrderTransverse(t.root);
	printf("\n");
	printf("%d\n", Height(t.root));
	for (index = 0; index < strlen(a); index++)
	{
		if (*(a + index) == '\0')                       //字符串结束就退出循环
			break;
		if (*(a + index) != '#')
			count++;
	}
	printf("%d ", count);
	count = 0;
	getLeavesCount(t.root, &count);
	printf("%d ", count);
	count = 0;
	OneChildCount(t.root, &count);
	printf("%d", count);
	getchar(); getchar();
	return 0;
}
void Input(char *a)
{
	char ch;
	int x = 0;
	while ((ch = getchar()) != '\n')
	{
		if (ch == ' ')
			continue;
		*(a + x) = ch;
		x++;
	}
	*(a + x) = '\0';                             //作为字符串结尾标志
}
BinaryTreeNode* PreCreateBinaryTree(BinaryTreeNode *t, char *a)
{
	char ch;
	t = (BinaryTreeNode*)malloc(sizeof(BinaryTreeNode));
	ch = *(a + i);
	i++;
	if (ch == '#')
		t = NULL;
	else
	{
		t->Data = ch;
		t->lChild = PreCreateBinaryTree(t->lChild, a);
		t->rChild = PreCreateBinaryTree(t->rChild, a);
	}
	return t;
}
void PreOrderTransverse(BinaryTreeNode *t)
{
	if (t != NULL)
	{
		printf(" %c", t->Data);
		PreOrderTransverse(t->lChild);
		PreOrderTransverse(t->rChild);
	}
}
void InOrderTransverse(BinaryTreeNode *t)
{
	if (t != NULL)
	{
		InOrderTransverse(t->lChild);
		printf(" %c", t->Data);
		InOrderTransverse(t->rChild);
	}
}
void PostOrderTransverse(BinaryTreeNode *t)
{
	if (t != NULL)
	{
		PostOrderTransverse(t->lChild);
		PostOrderTransverse(t->rChild);
		printf(" %c", t->Data);
	}
}
void getLeavesCount(BinaryTreeNode* T, int *count)
{
	if (T != NULL && T->lChild == NULL && T->rChild == NULL)
		*count = *count + 1;
	if (T)
	{
		getLeavesCount(T->lChild, count);
		getLeavesCount(T->rChild, count);
	}
}
void OneChildCount(BinaryTreeNode* T, int *count)
{
	if (T != NULL && ((T->lChild == NULL && T->rChild != NULL) || (T->lChild != NULL && T->rChild == NULL)))
		*count = *count + 1;
	if (T)
	{
		OneChildCount(T->lChild, count);
		OneChildCount(T->rChild, count);
	}
}

第二题题目：

多项式加法 1000ms 65536K
Description:
线性表是一种最简单、最基本，也是最常用的数据结构，其用途十分广泛，例如，用带表头结点的单链表求解一元整系数多项式加法和乘法运算。

现给两个一元整系数多项式，请求解两者之和。

Input:
两组数据，每一组代表一个一元整系数多项式，有多行组成，其中每一行给出多项式每一项的系数和指数，这些行按指数递减次序排序，每一组结束行为

0 -1

Output: 三组数据，前两组为一元整系数多项式，最后一组为两个多项式的和。

一元整系数多项式输出形式如下：

（1）多项式项4x输出为4X

（2）多项式项4x2输出为4X^2

（3）第一项系数为正数时，加号不要输出

（4）除常系数项外，项系数为1不显式输出，-1输出为-

例如，4x3- x2+x-1正确输出形式为4X^3-X2+X-1，错误输出形式为 +4X^3-1X2+1X-1
Sample Input:
3 14
-8 8
6 2
2 0
0 -1
2 10
4 8
-6 2
0 -1
Sample Output:
3X^14 - 8X^8 + 6X^2 + 2
2X^10 + 4X^8 - 6X^2
3X^14 + 2X^10 - 4X^8 + 2

第二题代码：

#include<stdio.h>

typedef struct pNode {
	int coef;                   //系数
	int exp;                    //指数
	struct pNode* link;         //连接下一项的指针
}PNode;
typedef struct polynominal {
	PNode *head;
}Polynominal;                   //带表头结点的单链表

void Init(Polynominal* p)
{
	p->head = (PNode*)malloc(sizeof(PNode));
	p->head->coef = 0;
	p->head->exp = -1;
	p->head->link = p->head;                          //单循环链表
}
void Create(Polynominal* p)
{
	PNode *pn, *pre, *q;
	Init(p);
	while (1)
	{
		pn = (PNode*)malloc(sizeof(PNode));           //循环每次开始时初始化pn
		scanf_s("%d%d", &pn->coef, &pn->exp);
		if (pn->exp == -1)
			break;
		if (pn->coef == 0)
			continue;
		pre = p->head;
		q = pre->link;
		while (q && q->exp > pn->exp)                   //多项式按指数从大到小排列
		{
			pre = q;
			q = q->link;
		}
		pn->link = q;
		pre->link = pn;                               //插入pn到单链表中
	}
}
void Copy(PNode* a, PNode* b)
{
	a->coef =  b->coef;
	a->exp = b->exp;
	a->link = b->link;
}

Polynominal* Add(Polynominal* p1, Polynominal* p2)
{
	Polynominal* p;                                                //作为输出
	PNode *pn1, *pn2,*pn,*temp;
	p = (Polynominal*)malloc(sizeof(Polynominal));
	Init(p);
	temp = (PNode*)malloc(sizeof(PNode));
	pn = p->head;
	pn1 = p1->head->link;
	pn2 = p2->head->link;
	while (!(pn1->exp == -1 && pn2->exp == -1))
	{
		temp = (PNode*)malloc(sizeof(PNode));
		if (pn1->exp > pn2->exp)
		{
			Copy(temp, pn1);
			temp->link = pn->link;
			pn->link = temp;
			pn1 = pn1->link;
			pn = pn->link;
		}
		else if (pn1->exp < pn2->exp)
		{
			Copy(temp, pn2);
			temp->link = pn->link;
			pn->link = temp;
			pn2 = pn2->link;
			pn = pn->link;
		}
		else
		{
			temp->coef = pn1->coef + pn2->coef;
			temp->exp = pn1->exp;
			temp->link = pn->link;
			pn->link = temp;
			pn1 = pn1->link;
			pn2 = pn2->link;
			pn = pn->link;
		}
	}
	return p;
}

void Output(Polynominal* p)
{
	PNode *pn;
	pn = p->head->link;                          //第一个数据结点
	if (pn->exp != 1 && pn->exp != 0 && pn->coef != 1 && pn->coef != 0)
		printf("%dX^%d", pn->coef, pn->exp);
	else if (pn->exp != 1 && pn->exp != 0 && pn->coef == 1)
		printf("X^%d", pn->exp);
	else if (pn->exp == 0 && pn->coef != 0)
		printf("%d", pn->coef);
	else if (pn->exp == 1 && pn->coef != 1 && pn->coef != 0)
		printf("%dX", pn->coef);
	else if (pn->exp == 1 && pn->coef == 1)
		printf("X");
	else if (pn->exp < 0)
		return;
	pn = pn->link;
	while (pn->exp >= 0)
	{
		if (pn->coef > 0 && pn->coef != 1 && pn->exp != 1 && pn->exp != 0)
			printf("+%dX^%d", pn->coef, pn->exp);
		else if (pn->coef > 0 && pn->coef != 1 && pn->exp == 1)
			printf("+%dX", pn->coef);
		else if (pn->coef > 0 && pn->exp == 0)
			printf("+%d", pn->coef);
		else if (pn->coef < 0 && pn->coef != -1 && pn->exp != 1 && pn->exp != 0)
			printf("%dX^%d", pn->coef, pn->exp);
		else if (pn->coef < 0 && pn->coef != -1 && pn->exp == 1)
			printf("%dX", pn->coef);
		else if (pn->coef < 0 && pn->exp == 0)
			printf("%d", pn->coef);
		else if (pn->coef == 1 && pn->exp != 1 && pn->exp != 0)
			printf("+X^%d", pn->exp);
		else if (pn->coef == -1 && pn->exp != 1 && pn->exp != 0)
			printf("X^%d", pn->exp);
		else if (pn->coef == 1 && pn->exp == 1)
			printf("+X");
		else if (pn->coef == -1 && pn->exp == 1)
			printf("-X");
		pn = pn->link;
	}
}

void main()
{
	Polynominal *p1, *p2,*p;
	p1 = (Polynominal*)malloc(sizeof(Polynominal));
	p2 = (Polynominal*)malloc(sizeof(Polynominal));
	Create(p1);
	Create(p2);
	p = Add(p1, p2);
	Output(p1);
	printf("\n");
	Output(p2);
	printf("\n");
	Output(p);
	getchar(); getchar();
}

第三题和第二题差不多，这里我就不多写了。。。
真心力交瘁
谢谢看到这里的大大大大大佬！！！
拜托帮忙解答一个问题也行！！！
谢谢！