【PAT】1020 Tree Traversals (25)（25 分）

1020 Tree Traversals (25)（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 (<=30), 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

C++代码如下：

 #include<iostream>
 #include<queue>
 using namespace std;

 struct node {
     int data;
     node *lchild, *rchild;
 };

 int post[], in[];
 int n;

 node* create(int posL,int posR,int inL,int inR) {
     if (posL>posR) return NULL;
     node *root = new node;
     root->data = post[posR];
     int k;
     for (k = inL; k <=inR; k++) {
         if (in[k] == post[posR]) break;
     }
     int numL = k-inL;
     root->lchild = create(posL, posL + numL-, inL,  k - );
     root->rchild = create(posL + numL, posR - , k + , inR);

     return root;
 }

 int num = ;

 void level(node*r) {
     if (r == NULL) return;
     queue<node*>q;
     q.push(r);
     while (!q.empty()) {
         node *top = q.front();
         q.pop();
         num++;
         cout << top->data;
         if (num < n) cout << ' ';
         if (top->lchild != NULL) q.push(top->lchild);
         if (top->rchild != NULL) q.push(top->rchild);
     }
 }

 int main() {
     cin >> n;
     int t;
     for (int i = ; i < n; i++) {
         cin >> t;
         post[i] = t;
     }
     for (int i = ; i < n; i++) {
         cin >> t;
         in[i] = t;
     }
     node*root = create(, n - , , n - );
     level(root);
     return ;
 }