PAT1127:ZigZagging on a Tree

1127. ZigZagging on a Tree (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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.

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

思路

类似Pat1029，根据中序遍历和后序遍历序列确定树，以层次遍历的形式存入vector<vector<int>>中，然后按每一层往返输出就行。

代码

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
vector<int> postorder(31);
vector<int> inorder(31);
vector<vector<int>> levels(31);


void buildTree(const int pl,const int pr,const int il,const int ir,const int level)
{

    if(pl > pr || il > ir)
        return;
    int root = postorder[pr],i = 0;
    while( inorder[il + i ] !=  root) i++;
    levels[level].push_back(root);
    buildTree(pl,pl + i - 1,il,il + i - 1,level + 1);
    buildTree(pl + i ,pr - 1,il + i + 1,ir,level + 1);
}


void zigzag()
{
    cout << levels[0][0];
    bool zigzag = false;
    for(int i = 1; i < levels.size() && !levels[i].empty();i++)
    {
        if(zigzag)
        {
            for(int j = levels[i].size() - 1;j >= 0;j--)
            {
                cout << " " << levels[i][j];
            }
        }
        else
        {
           for(int j = 0;j < levels[i].size();j++)
           {
               cout << " " << levels[i][j];
           }
        }
        zigzag = !zigzag;
    }
}


int main()
{
    int N;
    while(cin >> N)
    {
       for(int i = 0;i < N;i++)
         cin >> inorder[i];
       for(int i = 0;i < N;i++)
         cin >> postorder[i];
       buildTree(0,N - 1,0, N - 1,0);
       zigzag();
    }
}