1020 Tree Traversals

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

代码长度限制
16 KB
时间限制
400 ms
内存限制
64 MB

代码实现：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <queue>

using namespace std;

const int N = 40;

int n,count1;
int postorder[N], inorder[N];               //后序遍历，中序遍历
unordered_map<int, int> l, r, pos;          //用哈希表模拟二叉树

int build(int il, int ir, int pl, int pr)
{
	if(il>ir||pl>pr) return -1; 
    int root = postorder[pr];
    int k = pos[root];                      //得到根节点在中序遍历中的下标

    //k大于il表示根节点左边还有节点，即当前根节点存在左子树，下同
    //下面两行是难点，举例解释见图
	if(il<k)l[root] = build(il, k - 1, pl, pl + k - 1 - il);
	if(ir>k)r[root] = build(k + 1, ir, pl + k - il, pr - 1);

    return root;
}

void bfs(int root)                          //BFS用来层序遍历输出
{
    queue<int> q;
    q.push(root);
    while (q.size())
    {
        auto t = q.front();
        q.pop();
        if(t==-1)continue;
        if(count1==0)cout<<t;
        else cout <<" "<<t;
        count1++;
        if (l[t]) q.push(l[t]);       //判断该节点的左右儿子是否存在
        if (r[t]) q.push(r[t]);       //存在则加入队列，等待下一层遍历
    }
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++ ) cin >> postorder[i];
    for (int i = 0; i < n; i ++ )
    {
        cin >> inorder[i];
        pos[inorder[i]] = i;                //记录中序遍历每个点位置（剪枝）
    }

    int root = build(0, n - 1, 0, n - 1);   //参数为中序遍历区间和后序遍历区间
    bfs(root);

    return 0;
}