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[35],in[35];
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-1];
    h->v=mid;
    int i;
    for(i=0;i<n;i++){
        if(in[i]==mid){
            break;
        }
    }
    BuildTree(post,in,i,h->l);
    BuildTree(post+i,in+i+1,n-1-i,h->r);
}
int main(){
    int n,i;
    node *h;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%d",&post[i]);
    }
    for(i=0;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 0;
}