PAT 甲级 1020. Tree Traversals (25)

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

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<queue>
#include <cmath>
#include <map>
using namespace std;

vector<int> post, in, level(100000, -1);

void pre(int root, int start, int end, int index) {
	if(start > end) return;
	int i = start;
	while (i < end&&in[i] != post[root]) i++;
	level[index] = post[root];
	pre(root - 1 - end + i, start, i - 1, 2 * index + 1);
	pre(root - 1, i + 1, end, 2 * index + 2);
}
int main() {
	int n, cnt = 0;
	scanf("%d", &n);
	post.resize(n);
	in.resize(n);
	for (int i = 0; i < n; i++)
		scanf("%d", &post[i]);
	for (int i = 0; i < n; i++) {
		scanf("%d", &in[i]);
	}
	pre(n - 1, 0, n - 1, 0);
	for (int i = 0; i < level.size(); i++) {
		if (level[i] != -1 && cnt != n - 1) {
			printf("%d ", level[i]);
			cnt++;
		}
		else if (level[i] != -1) {
			printf("%d", level[i]);
			break;
		}
	}
	return 0;
}