PAT (Advanced Level) Practise 1020 Tree Traversals (25)

最新推荐文章于 2017-09-01 15:21:08 发布

Wang_128

最新推荐文章于 2017-09-01 15:21:08 发布

阅读量271

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT/PTA甲级 ----二叉树 ----dfs

本文链接：https://blog.youkuaiyun.com/a664607530/article/details/75579471

PAT/PTA甲级同时被 3 个专栏收录

174 篇文章

订阅专栏

----dfs

75 篇文章

订阅专栏

----二叉树

16 篇文章

订阅专栏

本文介绍了一种根据给定的后序遍历和中序遍历序列来构造并输出对应的二叉树层序遍历的方法。通过递归地构建二叉树结构，并使用队列进行层序遍历，实现了从输入序列到最终输出序列的有效转换。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题意：给你一棵树的后序遍历和中序遍历，输出这棵树的层序遍历

#include<queue>    
#include<cmath>    
#include<cstdio>    
#include<string>    
#include<cstring>    
#include<iostream>    
#include<algorithm>    
    
using namespace std;    
    
const int maxn = 50;    
int n, a[maxn], b[maxn], c[maxn][2];    
    
int dfs(int l,int r,int ll,int rr)    
{    
    if(r<l||rr<ll) return 0;    
    for(int i=ll;i<=rr;i++)    
    {    
        if(a[r]==b[i])    
        {    
            c[a[r]][0]=dfs(l,l+i-ll-1,ll,i-1);    
            c[a[r]][1]=dfs(l+i-ll,r-1,i+1,rr);    
        }    
    }    
    return a[r];    
}    
    
int main()    
{    
    while(~scanf("%d",&n))    
    {    
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);    
        for(int i=1;i<=n;i++) scanf("%d",&b[i]);    
        dfs(1,n,1,n);    
        queue<int>q;    
        q.push(a[n]);    
        int flag=0;    
        while(!q.empty())    
        {    
            if(flag) printf(" ");    
            else flag=1;    
            int e=q.front();    
            q.pop();    
            printf("%d",e);    
            if(c[e][0]) q.push(c[e][0]);    
            if(c[e][1]) q.push(c[e][1]);    
        }    
        printf("\n");    
    }    
    return 0;    
}