Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences, or preorder and inorder traversal sequences. However, if only the postorder and preorder traversal sequences are given, the corresponding tree may no longer be unique.

Now given a pair of postorder and preorder traversal sequences, you are supposed to output the corresponding inorder traversal sequence of the tree. If the tree is not unique, simply output any one of them.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 30), the total number of nodes in the binary tree. The second line gives the preorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first printf in a line Yes if the tree is unique, or No if not. Then print in the next line the inorder traversal sequence of the corresponding binary tree. If the solution is not unique, any answer would do. It is guaranteed that at least one solution exists. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input 1:

71 2 3 4 6 7 52 6 7 4 5 3 1

Sample Output 1:

Yes2 1 6 4 7 3 5

Sample Input 2:

41 2 3 42 4 3 1

Sample Output 2:

No2 1 3 4

1 #include 
     
       2 #include 
      
        3 using namespace std; 4 int n, a; 5 vector
       
        preOrder, postOrder, inOrder; 6 bool flag = true;//表示树的形态不唯一 7 void getInOrder(int root, int left, int right) 8 { 9     if (left >= right)10     {11         if (left == right)//只有一个节点12             inOrder.push_back(preOrder[root]);13         return;14     }15     int i = left;16     while (i < right && preOrder[root + 1] != postOrder[i])//查找前序遍历中下一个节点在后序中的位置17         ++i;18     if (i == right - 1)//先根序列中根节点的下一结点在后根序列中的位置正好等于right-119         flag = false;20     getInOrder(root + 1, left, i);21     inOrder.push_back(preOrder[root]);22     getInOrder(root + i - left + 2, i + 1, right - 1);23 }24 int main()25 {26     cin >> n;27     for (int i = 0; i < n; ++i)28     {29         cin >> a;30         preOrder.push_back(a);31     }32     for (int i = 0; i < n; ++i)33     {34         cin >> a;35         postOrder.push_back(a);36     }37     getInOrder(0, 0, n - 1);38     cout << (flag ? "Yes" : "No") << endl;39     for (int i = 0; i < n; ++i)40         cout << (i > 0 ? " " : "") << inOrder[i];41     cout << endl;42     return 0;43 }