给出先序序列，中序序列恢复二叉树

//给出先序序列，中序序列恢复出二叉树的方法

//方法1：利用递归的方法，首先先找出根节点，然后有先序和后序的特点可知：先序序列中在根节点左边的是左子树，根节点右面的是右子树

//有此来，进行递归，便可以恢复出二叉树

//方法2：方法2中栈的利用只是来存取，节点和先序中每个数据在后序的位置，利用pos 来记录每个位置的值

//该非递归的主要思想就是在后序序列中左节点的位置在右节点之前，也就是说pos(左)<pos(右)，先序序列是直接遍历下来就可以了，然后在

//后序序列中，查找每个数据的相对的位置的值，与前面的位置进行比较，来判断，该节点是左子树，还是右子树

//如果是左子树，则不断的将位置数据pos，和节点str不停的压入栈中

//如果是右子树，则将数据和节点不停的弹出栈，直到位置数据栈的栈顶元素大于当前位置数据的值，将该节点插入到该节点的右孩子节点处

//并将节点和位置数据pos全部压入栈中

//posChar(char ch,char* str)该函数是辅助函数，功能是查找ch在str中出现的位置

template<typename T>int BinaryTree<T>::posChar(char ch, char* str){if (strlen(str) == 0){cout << "the char* is empty" << endl;exit(true);}int i = 0;while (str[i] != '\0'){if (ch == str[i]){return i;}i++;}return -1;}template<typename T>void BinaryTree<T>::DefineBinaryTree(char* preOrder, char* midOrder){if (strlen(preOrder) == 0 || strlen(midOrder) == 0){this->root = NULL;return;}//记录每个与先序中节点对应的后序中的节点的位置int pos;//position,node两个栈存放的是pos,和生成的每个节点stack<int>position;stack<BinaryTreeNode<T>*>node;//str,ptr用于生成节点BinaryTreeNode<T>* str = NULL;BinaryTreeNode<T>* ptr = NULL;str = new BinaryTreeNode<T>(preOrder[0], NULL, NULL);this->root = str;pos = posChar(preOrder[0], midOrder);position.push(pos);node.push(str);int i = 1;while (preOrder[i] != '\0'){pos = posChar(preOrder[i], midOrder);if (pos == -1){cout << "the match is error" << endl;exit(true);}str = new BinaryTreeNode<T>(preOrder[i], NULL, NULL);if (pos < position.top()){//当num < position.top()时，则该节点是左节点ptr = node.top();ptr->leftChild = str;node.push(str);position.push(pos);}else{while (!position.empty() && pos > position.top()){ptr = node.top();position.pop();node.pop();}ptr->rightChild = str;node.push(str);position.push(pos);}i++;}}template<typename T>void BinaryTree<T>::simulate(){char preOrder[100];char midOrder[100];cout << "input the preOrder: ";cin >> preOrder;cout << "input the midOrder: ";cin >> midOrder;int length = strlen(midOrder);/*this->DefineBinaryTree(this->root,preOrder,midOrder,length);*/this->DefineBinaryTree(preOrder, midOrder);}template<typename T>void BinaryTree<T>::DefineBinaryTree(BinaryTreeNode<T>* &str, char* preOrder, char* midOrder, int length){if (length == 0){str = NULL;return;}str = new BinaryTreeNode<T>(*preOrder, NULL, NULL);char* pStr = strchr(midOrder, str->data);if (pStr == NULL){cout << "the midOrder string is wrong" << endl;exit(true);}int leftTreeLength = strlen(midOrder) - strlen(pStr);int rightTreeLength = length - leftTreeLength - 1;DefineBinaryTree(str->leftChild, preOrder + 1, midOrder, leftTreeLength);DefineBinaryTree(str->rightChild, preOrder + leftTreeLength + 1, pStr + 1, rightTreeLength);}