1127 ZigZagging on a Tree (30分)

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in “zigzagging order” – that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

zigzag.jpg
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 inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.
Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. 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:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

#include<bits/stdc++.h>

using namespace std;

const int N=40;

int n;
unordered_map<int,int> l,r,pos;
int in[N],post[N];
vector<int> res;

int build(int il, int ir, int pl, int pr)
{
    int root = post[pr];
    int k = pos[root];

    if (il < k) l[root] = build(il, k - 1, pl, pl + k - 1 - il);
    if (k < ir) r[root] = build(k + 1, ir, pl + k - 1 - il + 1, pr - 1);

    return root;
}

void bfs(int root)
{
    queue<int> q;
    q.push(root);
    bool flag=0;
    while(!q.empty())
    {
        int size=q.size();
        for(int j=0;j<size;j++)  //一层一层做bfs
        {
            auto i=q.front();
            q.pop();
            res.push_back(i);
            if(l[i]!=0) q.push(l[i]);
            if(r[i]!=0) q.push(r[i]);
        }
        flag=!flag;
        if(flag) reverse(res.begin()+res.size()-size,res.end()); //Z字形 这里做文章，一层反转一层不反转

    }
}
int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>in[i];
        pos[in[i]]=i;
    }
    for(int i=0;i<n;i++) cin>>post[i];

    int root=build(0,n-1,0,n-1);

    bfs(root);

    cout<<res[0];
    for(int i=1;i<res.size();i++)
        cout<<' '<<res[i];
    cout<<endl;
    return 0;
}