UVA 536 - Tree Recovery

最新推荐文章于 2020-03-10 16:22:37 发布

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#UVA #二叉树建树

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本文介绍了一种根据二叉树的先序遍历和中序遍历序列，重构二叉树并输出后序遍历的方法。通过分析输入字符串，使用递归算法构建树结构，最终实现树的遍历输出。

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二叉树先序遍历和中序遍历获得后序遍历

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:

/ \

B E

/ \ \

A C G

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!

Input

The input file 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.

Output

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

Sample Input

DBACEGF ABCDEFG BCAD CBAD

Sample Output

ACBFGED CDAB

#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
const int MAXN=10000;
char pre[MAXN];
char in[MAXN];
struct Node{
	char key;
	Node *right,*left;
};
Node *newnode (void)
{
	Node *T=(Node *)malloc(sizeof(Node));
	T->right=T->left=NULL;
	return T;
}
Node * Build(int l1,int r1,int l2,int r2,Node *T)
{
	T=newnode();
	T->key=pre[l1];
	int i;
	for (i=l2;i<=r2;i++)
	{
		if (in[i]==pre[l1])
			break;	
	}
	if (i+1<=r2)
	T->right=Build(r1-(r2-i)+1,r1,i+1,r2,T->right);
	if (i-1>=l2)
	T->left=Build(l1+1,r1-(r2-i),l2,i-1,T->left);
	return T;
}
void postTraversal(Node *T)
{
	if (T==NULL)
		return ;
	postTraversal(T->left);
	postTraversal(T->right);
	printf ("%c",T->key);
}
int main()
{
	int r1,r2,l1,l2;
	while (~scanf ("%s %s",pre,in))
	{
		Node *T=NULL;
		r1=r2=strlen(pre)-1;
		l1=l2=0;
		T=Build(l1,r1,l2,r2,T);
		postTraversal(T);
		memset(in,0,sizeof(in));
		memset (pre,0,sizeof(pre));
		printf ("\n");
	}
	return 0;
}