PAT (Advanced Level) Practise 1020 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

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
);
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;
}