Tree
时间限制：
1000 ms | 内存限制：
65535 KB
难度：
3
- 描述
-
Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes.
This is an example of one of her creations:
D / \ / \ B E / \ \ / \ \ A C G / / F
To record her trees for future generations, she wrote down two strings for each tree: a preorder traversal (root, left subtree, right subtree) and an inorder traversal (left subtree, root, right subtree). For the tree drawn above the preorder traversal is DBACEGF and the inorder traversal is ABCDEFG.
She thought that such a pair of strings would give enough information to reconstruct the tree later (but she never tried it).
Now, years later, looking again at the strings, she realized that reconstructing the trees was indeed possible, but only because she never had used the same letter twice in the same tree.
However, doing the reconstruction by hand, soon turned out to be tedious.
So now she asks you to write a program that does the job for her!
- 输入
-
The input will contain one or more test cases.
Each test case consists of one line containing two strings preord and inord, representing the preorder traversal and inorder traversal of a binary tree. Both strings consist of unique capital letters. (Thus they are not longer than 26 characters.)
Input is terminated by end of file. - 输出
- For each test case, recover Valentine's binary tree and print one line containing the tree's postorder traversal (left subtree, right subtree, root).
- 样例输入
-
DBACEGF ABCDEFG BCAD CBAD
- 样例输出
-
ACBFGED CDAB
- 上传者
#include<iostream>
#include<cstring>
#include<malloc.h>
using namespace std;
typedef struct Node{
Node * lchild,*rchild;
char value;
}Tree;
void ReBuild(char* PreOrder,char* InOrder,int TreeLen,Tree** root)
{
Tree *p;
char *LeftEnd;
if(PreOrder == NULL||InOrder == NULL||root == NULL)//检查边界条件
{
return;
}
p=(Tree *)malloc(sizeof(Tree));//获得前序遍历的第一个节点
p->value=*PreOrder;
p->lchild=p->rchild=NULL;
*root=p;
if(TreeLen==1)
return;//一个节点直接结束
LeftEnd=InOrder;//LeftEnd得到先序遍历根节点的值
while(*LeftEnd!=*PreOrder)//在中序遍历中查找该点的位置
{
LeftEnd++;
}
//寻找左子树长度
int LeftLen=0;
LeftLen=(int)(LeftEnd-InOrder);
//寻找右子树长度
int RightLen=0;
RightLen=TreeLen-LeftLen-1;
//重建左子树,递归
if(LeftLen>0)
ReBuild(PreOrder+1,InOrder,LeftLen,&(p->lchild));
//重建右子树，递归
if(RightLen>0)
ReBuild(PreOrder+LeftLen+1,InOrder+LeftLen+1,RightLen,&(p->rchild));
}
void PostOrder(Tree *p)//二叉树建完后进行后序遍历。
{
if(p!=NULL)
{
PostOrder(p->lchild);
PostOrder(p->rchild);
cout<<p->value;
}
}
int main()
{
char a[110],b[110];
Tree *p;
while(cin>>a>>b)
{
int len=strlen(a);
ReBuild(a,b,len,&p);
PostOrder(p);
cout<<endl;
}
return 0;
}