树——重建二叉树

①先处理传入的数组，如果数组没问题了，正式开始根据两个数组构建树

BitTree Construct(char *preorder, char *inorder, int len) {    if (preorder==nullptr || inorder==nullptr || len < 0)        return nullptr;    return ConstructCore(preorder, preorder + len - 1, inorder, inorder + len - 1);}

②传入先序遍历数组的首尾指针和中序遍历的首尾指针

BitTree ConstructCore(char *preorderbegin, char *preorderend, char *inorderbegin, char *inorderend) {    int rootValue = preorderbegin[0];    BitTree root = (BitTree)malloc(sizeof(BitNode));    root->data = rootValue;    root->left = nullptr;    root->right = nullptr;//构建树的节点    if (preorderbegin == preorderend) {        if (inorderbegin == inorderend&&*preorderbegin == *preorderend)//这里没搞懂①            return root;        else            throw std::exception("Invalid input");    }    char* rootInorder = inorderbegin;    while (rootInorder <= inorderend&&*rootInorder != rootValue)         ++rootInorder;    if(rootInorder==inorderend&&*rootInorder!=rootValue)        throw std::exception("Invalid input");    int leftlen = rootInorder - inorderbegin;    char *leftPreorderEnd = preorderbegin + leftlen;    if (leftlen > 0) {        root->left = ConstructCore(preorderbegin + 1, leftPreorderEnd, inorderbegin, rootInorder - 1);    }    if (leftlen < preorderend - preorderbegin)        root->right = ConstructCore(leftPreorderEnd + 1, preorderend, rootInorder + 1, inorderend);    return root;}