PAT 甲级 1020. Tree Traversals

原题传送门

树的遍历之已知后序和中序输出层序

#include <iostream>
#include <vector>
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 argc, const char * argv[]) {int n;cin >> n;post.resize(n);in.resize(n);for(int i = 0; i < n; ++i)cin >> post[i];for(int i = 0; i < n; ++i)cin >> in[i];pre(n-1, 0, n-1, 0);cout << level[0];for(int i = 1; i < level.size(); ++i) {if(level[i] != -1)cout << " " << level[i];}return 0;
}

附原题：

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

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