1127.ZigZagging on a Tree (30)

1127.ZigZagging on a Tree (30)

pat-al-1127

2017-03-04

今天考试把这题想错了……就得了1分……

写的有点麻烦，先按照层次遍历全部输出到数组，同时记录每层有多少个结点，输出的时候根据当前是奇数层还是偶数层来决定顺序还是逆序输出（由于保存了每层有多少个，可以把输出的下标范围限制住）

/**
* pat-al-1127
* 2017-03-04
* Cpp version
* Author: fengLian_s
*/
#include<stdio.h>
#include<queue>
#define MAX 35
using namespace std;
struct NODE
{
int key;
NODE* lchild;
NODE* rchild;
};
int in[MAX], post[MAX];
int n;
NODE* create(int inL, int inR, int postL, int postR)
{
if(postL > postR)
return NULL;
NODE *root = new NODE;
root->key = post[postR];
int k;
for(k = inL;k < inR;k++)
{
if(in[k] == post[postR])
break;
}
int len = k - inL;
root->lchild = create(inL, k-1, postL, postL+len-1);
root->rchild = create(k+1, inR, postL+len, postR-1);
return root;
}
int num[MAX], level[MAX], numOfThisLevel[MAX] = {1, 1}, nowLevel;
void bfs(NODE* root)
{
int front = 0, tail = 0;
queue<NODE*> q;
q.push(root);
level[0] = 1;
while(!q.empty())
{
NODE *tmp = new NODE;
tmp = q.front();
num[front] = tmp->key;
nowLevel = level[front++]+1;
q.pop();
if(tmp->lchild != NULL)
{
q.push(tmp->lchild);
level[++tail] = nowLevel;
numOfThisLevel[nowLevel]++;
}
if(tmp->rchild != NULL)
{
q.push(tmp->rchild);
level[++tail] = nowLevel;
numOfThisLevel[nowLevel]++;
}
}
}
void print()
{
int index = 0, l = 1, flagFirst = 1;
while(l <= nowLevel && index < n)
{
if(l % 2 == 0)//从左到右
{
//printf("l = %d\n", l);
for(int start = index;start < index + numOfThisLevel[l];start++)
{
//printf("start = %d\n", start);
if(flagFirst == 1)
flagFirst = 0;
else
printf(" ");
printf("%d", num[start]);
}
index += numOfThisLevel[l];
}
else
{
//printf("l = %d\n", l);
for(int start = index + numOfThisLevel[l] - 1;start >= index;start--)
{
//printf("start = %d: ", start);
if(flagFirst == 1)
flagFirst = 0;
else
printf(" ");
printf("%d", num[start]);
}
index += numOfThisLevel[l];
}
l++;
}
}
int main()
{
freopen("in.txt", "r", stdin);
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]);
}
NODE* root = create(0, n-1, 0, n-1);
bfs(root);
// for(int i = 0;i < n;i++)
// {
//   printf("%d ", num[i]);
// }
// for(int l = 1;l < nowLevel;l++)
// {
//   printf("level%d: %d\n", l, numOfThisLevel[l]);
// }
print();
}

-FIN-