PAT (Advanced Level) 1127. ZigZagging on a Tree (30) 解题报告

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：根据中序后序建树，然后层次输出，注意输出的顺序，题目要求Z型输出

代码：

#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
#define MAXN 1010000

int in[MAXN], post[MAXN], l[MAXN], r[MAXN];

int build(int inl, int inr, int postl, int postr)
{
if(inl > inr) return 0;
int rt = post[postr], p = inl, cnt;
while(rt != in[p]) p++;
cnt = p - inl;
l[rt] = build(inl, p-1, postl, postl + cnt - 1);
r[rt] = build(p+1, inr, postl + cnt, postr-1);
return rt;
}

void print(int rt)
{
queue<int> Q;
int level = 0;
Q.push(rt);
printf("%d", rt);
while(!Q.empty())
{
vector<int> V;
while(!Q.empty())
{
int now = Q.front();
Q.pop();
if(l[now]) V.push_back(l[now]);
if(r[now]) V.push_back(r[now]);
}
for(int i = 0; i < V.size(); i++)
Q.push(V[i]);
if(level % 2 == 0)
for(int i = 0; i < V.size(); i++)
printf(" %d", V[i]);
else
for(int i = V.size() - 1; i >= 0; i--)
printf(" %d", V[i]);
level ++;
}
printf("\n");
}

int main()
{
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++) scanf("%d", &in[i]);
for(int i = 0; i < n; i++) scanf("%d", &post[i]);
build(0, n-1, 0, n-1);
print(post[n-1]);
return 0;
}