pat甲1020. Tree Traversals（已知后序和中序求层次遍历）

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

tips:递归建树，广搜输出层次遍历

#include
#include

using namespace std;

int rt,n;
int t[320][2];//模拟建树
int post[320];//后序
int in[302];//中序
int flag;
queueq;
void build(int &x,int l,int r,int ll,int rr){
	if(l>r)return;
	x=post[r];
	
	int index;
	for(int i=ll;i<=rr;i++){
		if(in[i]==x){
			index=i;break;
		}
	}
	build(t[x][0],l,l+index-1-ll,ll,index-1);
	build(t[x][1],l+index-ll,r-1,index+1,rr);
	
}
void level_traversal(int x){
	q.push(x);
	while(!q.empty()){
		int tt=q.front();q.pop();
		printf("%s%d",flag==0?"":" ",tt);flag=1;
		
		if(t[tt][0])q.push(t[tt][0]);
		if(t[tt][1])q.push(t[tt][1]);
	}
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>post[i];
	for(int i=1;i<=n;i++)cin>>in[i];
	
	build(rt,1,n,1,n);
	
	//cout<