创建二叉树：输入0结束

void CreateTree(BinaryTree **pTree){    //输入数值    int nNum;    scanf("%d",&nNum);    if(nNum == 0)return;    *pTree = (BinaryTree*)malloc(sizeof(BinaryTree));    (*pTree)->nValue = nNum;    (*pTree)->pLeft = NULL;    (*pTree)->pRight = NULL;    //左    CreateTree(&((*pTree)->pLeft));    //右    CreateTree(&((*pTree)->pRight));}

创建二叉树：输入#结束，需要注意的是%c读入空格和换行，所以输入的时候要一连串输入，

void  CreateTree(Tree** tmp){    char num = 0;    scanf("%c",&num);    if(num == '#') return;    *tmp = (Tree*)malloc(sizeof(Tree));    (*tmp)->val = num - '0';    (*tmp)->pleft = NULL;    (*tmp)->pright = NULL;    CreateTree(&((*tmp)->pleft));    CreateTree(&((*tmp)->pright));}

测试数据：124##5##36###     

前序输出结果为：124536

函数为前序插入，只能为前序插入。因为必须确定根才能，建立左孩子和右孩子

将数组元素作为前序遍历的结果建立二叉树：

根从0开始，第2*i+1个节点为它的左孩子，2*i+2为它的右孩子

Tree* Create(int* arr,int len){    if(arr == NULL || len == 0) return NULL;    Tree* tree = (Tree*)malloc(sizeof(Tree)*len);        for(int i = 0; i < len; i++)    {            tree[i].val = arr[i];        if((2*i+1) < len)            tree[i].pleft = &tree[2*i+1];        else            (tree[i]).pleft = NULL;        if((2*i+2) < len)            tree[i].pright = &tree[2*i+2];        else            tree[i].pright = NULL;    }    return tree;}