pat甲级1020中序后序求层序

1020 Tree Traversals (25)（25 分）

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

已知后序中序求先序：

void pre(int root, int begin, int end)
{
    if (begin > end) return;
        cout  << post[root] << " "; 
    int p;
    for (p = begin; p <= end; p++)
    {
        if (in[p] == post[root])
            break;
    }
    pre(root - end + p - 1, begin, p - 1);
    pre(root - 1, p + 1, end);
}

后序中最后一个元素为跟，找出跟在中序中的位置就可以确定左右子树的节点个数，然后就可以递归的去求解。

这个题是求层序遍历，开始使用的构造树然后再广度优先搜索的方法，比较繁琐。后来看柳婼的博客发现使用数组的方法构造树比较方便。

若当前根的下标为i，则左节点下标为 2 * i + 1，右节点下标为 2 * i + 2。

#include <iostream>
#include <vector>
#include <math.h>
using namespace std;

vector<int> post, in, level(100000, -1);
void getLevel(int root, int begin, int end, int index);

int main()
{
    int N, i;
    cin >> N;
    post.resize(N);
    in.resize(N);
    for (i = 0; i < N; i++)
        cin >> post[i];
    for (i = 0; i < N; i++)
        cin >> in[i];
    getLevel(N - 1, 0, N - 1, 0);
    int cnt = 0;
    for (int i : level)
    {
        if (i != -1)
        {
            if (cnt > 0)
                cout << " ";
            cnt++;
            cout << i;
            if (cnt == N)
                break;
        }
    }
    return 0;
}

void getLevel(int root, int begin, int end, int index)
{
    if (begin > end) return;
    level[index] = post[root];
    int p;
    for (p = begin; p <= end; p++)
    {
        if (in[p] == post[root])
            break;
    }
    getLevel(root - end + p - 1, begin, p - 1, 2 * index + 1);
    getLevel(root - 1, p + 1, end, 2 * index + 2);
}