PAT甲级 1127. ZigZagging on a Tree (30)

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

—————————————————————————————————

题目的意思是给出一棵树的中序遍历和后序遍历，求奇数层反向输出偶数层正向输出的

层序遍历；

思路：先根据遍历建树，在层序输出奇偶分开讨论

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
int a[100005];
int b[100005];
struct node
{
    int v,l,r,deep;
} tree[100005];
int n,cnt;

int build(int l1,int r1,int l2,int r2,int deep)
{
    if(l1>r1||l2>r2) return -1;
    int ct=0;
    for(int i=l1; i<=r1; i++)
    {

        if(a[i]==b[r2])
        {
            break;
        }
        ct++;
    }
    int x=cnt;
    tree[cnt++].deep=deep;
    tree[x].v=b[r2];
    tree[x].l=build(l1,l1+ct-1,l2,l2+ct-1,deep+1);
    tree[x].r=build(l1+ct+1,r1,l2+ct,r2-1,deep+1);
    return x;
}

int main()
{
    scanf("%d",&n);
    for(int i=0; i<n; i++)
        scanf("%d",&a[i]);
    for(int i=0; i<n; i++)
        scanf("%d",&b[i]);
    cnt=0;
    int root=build(0,n-1,0,n-1,0);
    queue<node>q;
    stack<node>s;
    node f;
    q.push(tree[root]);
    int fl=0;
    while(!q.empty())
    {
        f=q.front();
        q.pop();
        if(f.l!=-1)
                q.push(tree[f.l]);
            if(f.r!=-1)
                q.push(tree[f.r]);
        if(f.deep%2==0)
        {
            s.push(f);

        }
        else
        {
            while(!s.empty())
            {
                if(fl++)
                    printf(" ");
                printf("%d",s.top().v);
                s.pop();
            }
            if(fl++)
                printf(" ");
            printf("%d",f.v);
        }
    }
        while(!s.empty())
            {
                if(fl++)
                    printf(" ");
                printf("%d",s.top().v);
                s.pop();
            }
    printf("\n");
    return 0;
}