题目：知道一棵二叉树的前序序列、中序系列，打印出它的后序系列

下面算法的时间复杂度：O(n^2) -- 最坏情况就是左单支树

/*    Kown the preorder listing and the inorder listing of a tree, print it's postorder listing*/void PrintPostOrder(const char *pre, int s1, int e1,                    const char *in,  int s2, int e2){    if (s1 > e1 || s2 > e2)        return;    char root = pre[s1];    for (int i=s2; i<=e2; i++)        if (in[i] == root)            break;    int offset = i-s2;    PrintPostOrder(pre, s1+1, s1+offset, in, s2, i-1);    PrintPostOrder(pre, s1+offset+1, e1, in, i+1, e2);    printf("%c", root);}int main(){    const char *pre="ABCDEF";    const char *in="FEDCBA";    printf("PreOrder : %s\n", pre);    printf("InOrder  : %s\n", in);    printf("PostOrder: ");    PrintPostOrder(pre, 0, strlen(pre)-1, in, 0, strlen(in)-1);    printf("\n");    return 0;}