pat

题目

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

由二叉树的后序遍历和中序遍历获得层序遍历的结果。注意最好是用pair或者结构体来存储递归过程中的层序遍历的结果。

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#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
vi post, in;
vii level;
void _Post(int root, int _start, int _end, int index){
    if(_start > _end) return;
    int i = _start;
    while(i < _end && in[i] != post[root])  i++;
    level.push_back(ii(index, in[i]));
    _Post(root-1-_end+i, _start, i-1, 2  index + 1);
    _Post(root-1, i+1, _end, 2  index + 2);
}
int (){
    int N;
    scanf("%d", &N);
    post.assign(N, 0); in.assign(N, 0);
    for(int i = 0;i < N;i++)    scanf("%d", &post[i]);
    for(int i = 0;i < N;i++)    scanf("%d", &in[i]);
    _Post(N-1, 0, N-1, 0);
    sort(level.begin(), level.end());
    for(int i = 0;i < level.size();i++){
        printf("%d", level[i].second);
        if(i != level.size()-1) printf(" ");
    }
    return 0;
}