nyoj 221 Tree 【已知前序中序求後序】



Tree

時間限制： 1000 ms  |  記憶體限制： 65535 KB 難度： 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! 

輸入

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.

輸出

For each test case, recover Valentine's binary tree and print one line containing the tree's postorder traversal (left subtree, right subtree, root).

樣例輸入

DBACEGF ABCDEFG
BCAD CBAD      

樣例輸出

ACBFGED
CDAB      

已知前序中序求後序，我認為它就是一個DFS。 看了好多非連結清單的求法，自己根據喜好寫了個便于自己了解的。      

#include <cstdio>
#include <cstring>
char preorder[30];
char inorder[30];
char postorder[30];
void build(int n, int pre, int in, int rec)
{
	if(n <= 0) return ;//無子樹 
	int i;
	for(i = 0; ; i++)
	{
		if(preorder[pre] == inorder[in+i])//找到位置 
		break;
	} 
	build(i, pre+1, in, rec);//左子樹 
	build(n-i-1, pre+i+1, in+i+1, rec+i);//右子樹 
	postorder[n-1+rec] = preorder[pre];//記錄 
}
int main()
{
	int n;
	while(scanf("%s%s", preorder, inorder) != EOF)
	{
		n = strlen(preorder);
		build(n, 0, 0, 0);
		postorder[n] = '\0'; 
		printf("%s\n", postorder);
	}
	return 0;
} 
           
   

附上一個劉汝佳的指針版本：      

#include <cstdio>
#include <cstring>
char preorder[30];
char inorder[30];
char postorder[30];
void build(int n, char *s1, char *s2, char *s3)
{
	if(n <= 0) return ;
	int pos = strchr(s2, s1[0]) - s2;//在中序序列中找根節點的位置
	build(pos, s1+1, s2, s3);//找左子樹的後序序列 
	build(n-pos-1, s1+pos+1, s2+pos+1, s3+pos);//找右子樹的後序序列 
	s3[n-1] = s1[0];//記錄 
}
int main()
{
	int n;
	while(scanf("%s%s", preorder, inorder) != EOF)
	{
		n = strlen(preorder);
		build(n, preorder, inorder, postorder);
		postorder[n] = '\0';
		printf("%s\n", postorder);
	} 
	return 0;
}