UVa 536 - Tree Recovery 前中序求后序

最新推荐文章于 2022-03-20 16:04:59 发布

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本博客介绍了一种利用给定的前序和中序遍历字符串来重建二叉树的方法，并展示了如何用C++实现这一过程。重点在于通过解析前序遍历获取根节点，然后利用中序遍历来划分左右子树，最终得到完整的二叉树结构。文章以一个具体的例子展示重建过程，并给出了完整的C++代码实现。

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Tree Recovery

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!

Input Specification

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 Specification

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 <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int MAX = 26+4;

struct Node{
	char elem;
	struct Node *left;
	struct Node *right;
};

void aftOrder(char *in, char *pre, int length)
{
	if(length==0)
		return ;
	Node node;
	node.elem = *pre;
	int rootIndex = 0;
	for(;rootIndex<length; rootIndex++)
	{
		if(in[rootIndex]== *pre)
			break;
	}
	aftOrder(in, pre+1, rootIndex);// 左子树
	aftOrder(in+rootIndex+1, pre+rootIndex+1,
		length-(rootIndex+1));// 右子树
	cout << node.elem;// 输出root
	return ;
}


int main()
{
//	freopen("in.txt","r",stdin);
	char pre[MAX], in[MAX];
	while(cin>>pre)
	{
		cin>>in;
		aftOrder(in, pre, strlen(pre));
		cout << endl;
		memset(pre, '\0', sizeof(pre));
		memset(in, '\0', sizeof(in));
	}
	
	return 0;
}