Tsinghua machine trial: There is a forward, middle, and back order output.

 

 

Description:

Binary Tree pre-order, middle-order, and post-order traversal definitions:
Forward traversal: For any subtree, first access the heel, then traverse its left subtree, and finally traverse its right subtree;
In-order traversal: For any subtree, first traverse the left subtree, then access the root, and finally traverse the right subtree;
Post-order traversal: For any subtree, first traverse the left subtree, then traverse the right subtree, and finally access the root.
Specify the pre-order traversal and Middle-order traversal of a binary tree, and obtain the post-order traversal (Note: given the pre-order traversal and Middle-order traversal, the post-order traversal can be uniquely determined ).

Input:

The length of N is equal to or less than 26.
First, first, and second.
The node names in the binary tree are represented by uppercase letters: A, B, C... a maximum of 26 nodes.

Output:

There may be multiple groups of input samples. For each group of test samples,
Output a row, which is a string that is traversed in the descending order.

Sample input:

ABCBACFDXEAGXDEFAG

Sample output:

BCAXEDGAF

Package Tsinghua;

Import java. Io. bufferedinputstream;
Import java. util. collections;

Public class evaluate the descending order traversal {
Static string str1, str2;
Public static void main (string [] ARGs ){
S = new second (New bufferedinputstream (system. In ));
While (S. hasnext ()){
Str1 = S. Next ();
Str2 = S. Next ();
Node root = creattree (str1, str2 );
Lastsee (Root );
System. Out. println ();
}

}

Static void lastsee (node tree ){
If (tree = NULL) return;
Lastsee (tree. L );
Lastsee (tree. R );
System. Out. Print (tree. data );
}
Static node creattree (string S1, string S2 ){
Node n = NULL;
If (S1 = NULL | S2 = NULL)
Return NULL;
String left2 = NULL, Right2 = NULL, news1 = NULL, news2 = NULL;
Int len1 = 0;
Char root = s1.charat (0 );
Int Index = s2.indexof (Root );
If (index> 0 ){
Left2 = s2.substring (0, index );
Len1 = left2.length ();
}
If (index <s2.length ()-1)
Right2 = s2.substring (index + 1 );
News1 = s1.substring (1, 1 + len1 );
News2 = s1.substring (1 + len1 );
N = new node (creattree (news1, left2), creattree (news2, Right2), s1.charat (0 ));
Return N;
}
}

Class node {
Node L;
Node R;
Char data;
Public node (node l, node R, char data ){
This. L = L;
This. r = R;
This. Data = data;
}

}

 

 

There are also methods that do not need to be built, copy others' code, and learn.

char a[30],b[30];
void Bice(int pf,int pr,int mp,int mr)
{
if(pf>pr)
return;
int i;
char c=a[pf];
for(i=0;i<=mr-mp;i++)
if(b[mp+i]==c)
break;
    Bice(pf+1,pf+i,mp,mp+i-1);
    Bice(pf+i+1,pr,mp+i+1,mr);
    cout<<c;
}
int main()
{
while(cin>>a>>b)
    {
int n=strlen(a);
        Bice(0,n-1,0,n-1);
        cout<<endl;
    }
}