1020. Tree Traversals (25)-蒲公英云

1020. Tree Traversals (25)

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

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 postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. 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:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

题目大意：

假设二叉树中所有的键值都是不同的正整数。给出后序遍历和中序遍历序列，你的任务是输出相应的层次遍历序列。
输入规格：
每个输入文件包含一个测试用例。对于每个测试用例，第一行给出一个正整数N（<=30）,即二叉树中的节点总数。第二行给出后序遍历序列，第三行给出中序遍历序列。直线上得所有数字都用空格隔开。
输出规范：
对于那个测试用例，在一行中打印出相应的层次遍历序列。直线上得所有数字用一个空格隔开，在这一行的末尾没有多余的空格。

代码：

#include<stdio.h>
#include<stdlib.h>
#include<queue>
using namespace std;
struct node
{
    int data;
    struct node *lchild,*rchild;
};
struct node *create(int n,int *a,int *b)
{
struct node *root;
int *p;
if(n==0)
return NULL;
root=(struct node *)malloc(sizeof(struct node));
root->data=a[n-1];
for(p=b;p<p+n;p++)
if(*p==a[n-1])
break;
int t;
t=p-b;
root->lchild=create(t,a,b);
root->rchild=create(n-t-1,a+t,p+1);
return root;
}
void level(struct node *T)
{
    int i=0;
    struct node *p;
    queue<struct node *> que;
    que.push(T);
    while(!que.empty())
    {
        if(que.front())
        {
            p=que.front();
        que.push(p->lchild);
        que.push(p->rchild);
       if(i==0)
       {
           i++;
           printf("%d",p->data);
       }
       else
       {
           printf(" %d",p->data);
       }
        }
        que.pop();
    }
}
void cengci(struct node *root)
{
int out=0,in=0,i=0;
struct node *q[100];
q[in++]=root;
while(in>out)
{
if(q[out])
{
    if(i==0)
{
    i++;
    printf("%d",q[out]->data);
}
else
   printf(" %d",q[out]->data);
q[in++]=q[out]->lchild;
q[in++]=q[out]->rchild;
}
out++;
}
}
int main()
{
    int i,j,n,m,k,t,a[31],b[31];
    struct node *root;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0;i<n;i++)
    {
        scanf("%d",&b[i]);
    }
    root=create(n,a,b);
    level(root);
    //cengci(root);
    return 0;
}