PAT 2020年秋 7-3 Left-View of Binary Tree (25 分) AC代码

7-3 Left-View of Binary Tree (25 分)

The left-view of a binary tree is a list of nodes obtained by looking at the tree from left hand side and from top down. For example, given a tree shown by the figure, its left-view is { 1, 2, 3, 4, 5 }

Given the inorder and preorder traversal sequences of a binary tree, you are supposed to output its left-view.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤20), which is the total number of nodes in the tree. Then given in the following 2 lines are the inorder and preorder traversal sequences of the tree, respectively. All the keys in the tree are distinct positive integers in the range of int.

Output Specification:

For each case, print in a line the left-view of the tree. All the numbers in a line are separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.

Sample Input:

8
2 3 1 5 4 7 8 6
1 2 3 6 7 4 5 8

结尾无空行

Sample Output:

1 2 3 4 5

结尾无空行

主要还是考察树的序列转换和遍历，输出层序遍历的第一个点就好了

AC代码：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int N;
vector<int> in,pre,le[20];
void trail(int root,int start,int end,int level){
	if(start>end) return;
	int index=pre[root];
	le[level].push_back(index);
	int i=start;
	while(in[i]!=index) i++;
	trail(root+1,start,i-1,level+1);
	trail(root+i-start+1,i+1,end,level+1);
}
int main(){
	cin>>N;
	in.resize(N);
	pre.resize(N);
	for(int i=0;i<N;i++){
		scanf("%d",&in[i]);
	}
	for(int i=0;i<N;i++){
		scanf("%d",&pre[i]);
	}
	trail(0,0,N-1,0);
	printf("%d",pre[0]);
	for(int i=1;i<N;i++){
		if(le[i].size()!=0){
			printf(" %d",le[i][0]);
		}else{
			cout<<endl;
			break;
		}
	}
	return 0;
}