【PAT】【Advanced Level】1020. Tree Traversals (25)

最新推荐文章于 2024-05-13 09:10:43 发布

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最新推荐文章于 2024-05-13 09:10:43 发布

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本文介绍了一种根据给定的后序和中序遍历序列来构建对应的二叉树，并输出该树的层序遍历序列的方法。通过递归地确定每个子树的根节点及其左右子树，最终实现从后序和中序到层序遍历的转换。

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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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1020

https://www.nowcoder.com/questionTerminal/b1dd2f978d9f49cdb936a5061e590837

思路：

DFS判断

对于一个子树

根节点肯定在后序序列中代表这个子树的区间的末尾

根据根节点在中序序列中寻找，分开左右子树，统计个数并回到后序序列找到子树的根节点

以此类推

最后按层序输出

坑点：

最后的统计或者不用排序，或者一定要用稳定的排序方式（冒泡等），快速排序不稳定

CODE:

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

int n;
int po[31];
int io[31];
typedef struct S
{
    int nu;
    int flo;
};
int flor=0;
S re[31];
int num=0;
int mm=0;
void fi(int ro,int l,int r,int l1,int r1)
{
    mm=max(mm,flor);
    //cout<<ro<<" "<<l<<" "<<r<<" "<<l1<<" "<<r1<<endl;
    //cout<<po[ro]<<endl;
    S t;
    t.nu=po[ro];
    t.flo=flor;
    re[num]=t;
    num++;
    if (l==r) return ;
    int pos=0;
    for (int i=l;i<=r;i++)
        if(io[i]==po[ro])
    {
        pos=i;
        break;
    }
    //cout<<pos<<endl;
    int nr=l1+pos-l-1;
    if(pos!=l)
    {
        //cout<<nr<<" "<<l<<" "<<pos-1<<" "<<l1<<" "<<nr-1<<endl;
        flor++;
        fi(nr,l,pos-1,l1,nr-1);
        flor--;
    }
    if (pos!=r)
    {
        //cout<<r1<<" "<<pos+1<<" "<<r<<" "<<nr+1<<" "<<r1-1<<endl;
        flor++;
        fi(r1,pos+1,r,nr+1,r1-1);
        flor--;
    }
    return ;
}


int main()
{
    cin>>n;
    for (int i=1;i<=n;i++)   cin>>po[i];
    for (int i=1;i<=n;i++)   cin>>io[i];
    fi(n,1,n,1,n-1);
    for (int i=num-1;i>0;i--)
        for (int j=0;j<i;j++)
    {
        if (re[j].flo>re[j+1].flo)
        {
            S t=re[j];
            re[j]=re[j+1];
            re[j+1]=t;
        }
    }
    for (int i=0;i<num-1;i++) cout<<re[i].nu<<" ";
    cout<<re[num-1].nu;
    /*
    int re1[31];
    int nn=0;
    for (int i=0;i<=mm;i++)
    for (int j=0;j<num;j++)
    {
        if (re[j].flo==i)
        {
            re1[nn]=re[j].nu;
            nn++;
        }
    }
    for (int i=0;i<nn-1;i++) cout<<re1[i]<<" ";
    cout<<re1[nn-1];
    */
    return 0;
}