【PAT】【Advanced Level】1127. ZigZagging on a Tree (30)

1127. ZigZagging on a Tree (30)

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. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in "zigzagging order" -- that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

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 inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. 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:

812 11 20 17 1 15 8 512 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

原题链接：

https://www.patest.cn/contests/pat-a-practise/1127

思路：

建树，然后层序排序，针对层数相同的，奇数层按中序遍历次序的降序排序，偶数层按中序遍历的升序排序。

CODE：

#include<iostream>#include<vector>#include<algorithm>#define N 31using namespace std;typedef struct S{int val;int ls;int rs;int fl;int tra;};vector<S> T;int in[N];int po[N];void dfs(int l1,int r1,int l2 ,int r2,int flo){if (l1==r1){S ne;ne.val=in[l1];ne.fl=flo;ne.tra=l1;T.push_back(ne);return ;}int pos;for(int i=l1;i<=r1;i++){if (in[i]==po[r2]){pos=i;break;}}int le1=pos-l1;int le2=r1-pos;if (pos!=l1){dfs(l1,pos-1,l2,l2+le1-1,flo+1);}S ne;ne.val=in[pos];ne.fl=flo;ne.tra=pos;T.push_back(ne);if (pos!=r1){dfs(pos+1,r1,l2+le1,r2-1,flo+1);}}bool cmp(S a, S b){if (a.fl==b.fl){if (a.fl%2==1){return a.tra>b.tra;}else{return a.tra<b.tra;}}else{return a.fl<b.fl;}}int main(){int n;cin>>n;for (int i=0;i<n;i++) cin>>in[i];for (int i=0;i<n;i++) cin>>po[i];dfs(0,n-1,0,n-1,1);sort(T.begin(),T.end(),cmp);for (int i=0;i<n;i++){if (i!=0) cout<<" ";cout<<T[i].val;}return 0;}