pat1020. Tree Traversals (25)

1020. Tree Traversals (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

---

提交代码

 #include<cstdio>
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<queue>
 #include<vector>
 #include<cmath>
 using namespace std;
 int post[],in[];
 struct node{
     int v;
     node *l,*r;
     node(){
         l=r=NULL;
     }
 };
 void BuildTree(int *post,int *in,int n,node *&h){
     if(!n){
         return;
     }
     h=new node();
     int mid=post[n-];
     h->v=mid;
     int i;
     for(i=;i<n;i++){
         if(in[i]==mid){
             break;
         }
     }
     BuildTree(post,in,i,h->l);
     BuildTree(post+i,in+i+,n--i,h->r);
 }
 int main(){
     int n,i;
     node *h;
     scanf("%d",&n);
     for(i=;i<n;i++){
         scanf("%d",&post[i]);
     }
     for(i=;i<n;i++){
         scanf("%d",&in[i]);
     }
     BuildTree(post,in,n,h);
     node cur;
     queue<node> q;
     q.push(*h);
     printf("%d",h->v);
     while(!q.empty()){
         cur=q.front();
         q.pop();
         if(cur.l!=NULL){
             q.push(*(cur.l));
             printf(" %d",cur.l->v);
         }
         if(cur.r!=NULL){
             q.push(*(cur.r));
             printf(" %d",cur.r->v);
         }
     }
     return ;
 }