二叉树已知（中序+一序）求另一序

#include <iostream>#include <cstring>using namespace std;char pre[50] = "ABDHLEKCFG";        //前序序列char mid[50] = "HLDBEKAFCG";        //中序序列char post[50] = "LHDKEBFGCA";        //后序序列typedef struct _Node{    char v;    struct _Node *left;    struct _Node *right;}Node, *PNode;void PostTravelTree(PNode pn);        //树的后序递归遍历void PreTravelTree(PNode pn);        //树的前序递归遍历void PreMidCreateTree(PNode &pn, int i, int j, int len);        //利用前序中序序列创建树void PostMidCreateTree(PNode &pn, int i, int j, int len);        //利用后序中序序列创建树int Position(char c);                //确定c在中序序列mid中的下标,假设树的各个节点的值各不相同int main() {     PNode root1 = NULL, root2= NULL;    PreMidCreateTree(root1, 0, 0, strlen(mid));    PostTravelTree(root1);    cout<<endl;    PostMidCreateTree(root2, strlen(post)-1, 0, strlen(mid));    PreTravelTree(root2);    cout<<endl;    return 0;}int Position(char c){    cout<<strchr(mid,c)<<endl;    return strchr(mid,c)-mid;}/*  利用前序中序序列创建树,参考了 *        i: 子树的前序序列字符串的首字符在pre[]中的下标 *        j: 子树的中序序列字符串的首字符在mid[]中的下标 *      len: 子树的字符串序列的长度 */void PreMidCreateTree(PNode &pn, int i, int j, int len){    if(len <= 0){        pn = NULL;        return;    }    pn = new Node;    pn->v = pre[i];    int m = Position(pre[i]);    PreMidCreateTree(pn->left, i+1, j, m-j);            //m-j为左子树字符串长度    PreMidCreateTree(pn->right, i+(m-j)+1, m+1, len-1-(m-j));    //len-1-(m-j)为右子树字符串长度}/*  利用后序中序序列创建树 *        i: 子树的后序序列字符串的尾字符在post[]中的下标 *        j: 子树的中序序列字符串的首字符在mid[]中的下标 *      len: 子树的字符串序列的长度 */void PostMidCreateTree(PNode &pn, int i, int j, int len){    if(len <= 0){        pn = NULL;        return;    }    pn = new Node;    pn->v = post[i];    int m = Position(post[i]);    PostMidCreateTree(pn->left, i-1-(len-1-(m-j)), j, m-j);//注意参数:m-j左子树的长度,len-1-(m-j)右子树的长度    PostMidCreateTree(pn->right, i-1, m+1, len-1-(m-j));}void PostTravelTree(PNode pn)        //后序递归遍历{    if(pn)    {        PostTravelTree(pn->left);            PostTravelTree(pn->right);        cout<<pn->v<<" ";    }}void PreTravelTree(PNode pn)        //前序递归遍历{    if(pn)    {        cout<<pn->v<<" ";        PreTravelTree(pn->left);            PreTravelTree(pn->right);    }}