Binary Tree Traversals

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3475    Accepted Submission(s): 1555

Problem Description

A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary tree can be systematically traversed or ordered. They are preorder, inorder and postorder. Let T be a binary tree with root r and subtrees T1,T2. In a preorder traversal of the vertices of T, we visit the root r followed by visiting the vertices of T1 in preorder, then the vertices of T2 in preorder. In an inorder traversal of the vertices of T, we visit the vertices of T1 in inorder, then the root r, followed by the vertices of T2 in inorder. In a postorder traversal of the vertices of T, we visit the vertices of T1 in postorder, then the vertices of T2 in postorder and finally we visit r. Now you are given the preorder sequence and inorder sequence of a certain binary tree. Try to find out its postorder sequence.

Input

The input contains several test cases. The first line of each test case contains a single integer n (1<=n<=1000), the number of vertices of the binary tree. Followed by two lines, respectively indicating the preorder sequence and inorder sequence. You can assume they are always correspond to a exclusive binary tree.

Output

For each test case print a single line specifying the corresponding postorder sequence.

Sample Input

9 1 2 4 7 3 5 8 9 6 4 7 2 1 8 5 9 3 6

Sample Output

7 4 2 8 9 5 6 3 1

Source

HDU 2007-Spring Programming Contest

给定一课二叉树的先序和中序，求这课数的后序：

      采用划区域的方法，逐步缩小区间求解

  分析：

代码：

 1 /*hdu 1710 二叉树*/
 2 //#define LOCAL
 3 #include<cstdio>
 4 #include<cstring>
 5 using namespace std;
 6 const int maxn=1005;
 7 int aa[maxn],bb[maxn];
 8 void dfs(int a,int b,int n,int tag)
 9 {
10     int i;
11     if(n<=0)return ;
12     for(i=0;aa[a]!=bb[b+i];i++);
13     dfs(a+1,b,i,0);
14     dfs(a+i+1,b+i+1,n-i-1,0);
15     printf("%d",aa[a]);
16     if(!tag)printf(" ");
17 }
18 int main()
19 {
20   #ifdef LOCAL
21     freopen("test.in","r",stdin);
22   #endif
23   int n,i,k,pre;
24   while(scanf("%d",&n)!=EOF)
25   {
26       for(i=1;i<=n;i++)
27       scanf("%d",aa+i);
28     for(i=1;i<=n;i++)
29       scanf("%d",bb+i);
30     dfs(1,1,n,1);
31     printf("n");
32   }
33  return 0;
34 }

 顺便做了一个二叉树知道其中两个序列求第三个序列的模板吧！  注意： 知道前序和后序是无法求出唯一二叉树的！

代码：

//这部分检验aa[]为先序,bb[]为中序
void dfs_1(char aa[],char bb[], int a,int b,int n)
{
    if(n<=0)return ;
    int i;
    for(i=0;aa[a]!=bb[b+i];i++);
    dfs_1(aa,bb,a+1,b,i);     //左子树
    dfs_1(aa,bb,a+i+1,b+i+1,n-i-1);     //右子树
    printf("%c",aa[a]);
}
  //aa[]为中序，bb[]为后序
void dfs_3(char aa[],char bb[],int a,int b,int n)
{
       if(n<=0)return ;
      printf("%c",bb[b]);
     int i;
     for(i=0;bb[b]!=aa[a+i];i++);
     dfs_3(aa,bb,a,b-n+i,i);     //左子树
     dfs_3(aa,bb,a+i+1,b-1,n-i-1); //右子树
}