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:

72 3 1 5 7 6 41 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;}