1020 Tree Traversals (25 分)

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

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

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. 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:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2
分析：根据后序遍历和中序遍历 输出层序遍历
我的做法是利用后序遍历和中序遍历建树 再层序遍历输出
利用两个变量来记录我们需要建子树的范围 另一个变量记录每层的根节点

 #define _CRT_SECURE_NO_WARNINGS
 #include<iostream>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<algorithm>
 using namespace std;
 typedef struct BtNode* Bt;
 struct BtNode
 {
     int Num;
     Bt RT;
     Bt LT;
 };
 vector<vector<int> > Order();
 Bt BuildTree(Bt T,int i2,int j2,int j1)
 {
     if (i2 >= j2)
         return T;
     int i = i2;
     for (; i < j2; i++)
     {
         if (Order[][i] == Order[][j1 - ])
             break;
     }
     if (!T)
     {
         T = new BtNode();
         T->LT = NULL;
         T->RT = NULL;
         T->Num = Order[][i];
     }
     T->LT =BuildTree(T->LT,i2,i,j1-j2+i);
     T->RT = BuildTree(T->RT,i+,j2,j1-);
     return T;
 }
 int main()
 {
     int N;
     cin >> N;
     int num;
     for(int i=;i<;i++)
         for (int j = ; j < N; j++)
         {
             cin >> num;
             Order[i].push_back(num);
         }
     Bt T=NULL;
     T = BuildTree(T,,N,N);
     queue<Bt> Q;
     Q.push(T);
     cout << T->Num;
     while (!Q.empty())
     {
         if (Q.front()->LT)
         {
             Q.push(Q.front()->LT);
             cout << " " << Q.front()->LT->Num;
         }
         if (Q.front()->RT)
         {
             Q.push(Q.front()->RT);
             cout << " " << Q.front()->RT->Num;
         }
         Q.pop();
     }
     return ;
 }