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