[HOJ]1452 Tree Recovery(对二叉树的先序、中序、后序的熟悉与掌握)

这个题主要是由给定的二叉树的先序和中序序列来还原二叉树，并且能再给出后序序列，重点和难点都是在如何还原二叉树，后序遍历只是检测是否还原正确的一个手段而已；

由于先序序列的特点是先访问根节点在访问左右节点，中序序列则是先左节点->根节点->右节点；

所以我们可以先由先序序列得到根，再到中序序列找到根的位置，而根的左边的就是其左子树的节点，右边的就是其右子树的节点，依次，递归下去就能还原二叉树，不过在递归的过程中应考虑到左左、左右、右左、右右等几种特殊的情况，有一个好的判别方法；

其他就是，还会有部分关于字符串的处理，需要掌握相应的知识。

题目: 

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

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.

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 ABCDEFGBCAD CBAD

Sample Output

ACBFGEDCDAB

#include <iostream>#include <cstdlib>#include <string>using namespace std;struct node{    node *left;    node *right;    char data;};string str,porderStr,inorderStr;int n = 0;node* GetLeftRoot(string str1);node* GetRightRoot(string str2);void BuiltTree(node *root,int num);void postOrderBT(node *root);node* GetLeftRoot(string str1){    string leftStr,rightStr;    int position = -1;    node *root,*leftree,*rightree;    char ch;    int length = 0;    ch = porderStr[n];    n++;    position = str1.find(ch);    root = new node;    root->data = ch;    length = str1.length();    if(position == 0){        if(length > 1){            rightStr.assign(str1,1,length);            rightree = GetRightRoot(rightStr);            root->left = NULL;            root->right = rightree;            return root;        }        else{            root->left = NULL;            root->right = NULL;            return root;        }    }    else{        leftStr.assign(str1,0,position);        rightStr.assign(str1,position+1,length);        if(!leftStr.empty()){            leftree = GetLeftRoot(leftStr);            root->left = leftree;        }        else            root->left = NULL;        if(!rightStr.empty()){            rightree = GetRightRoot(rightStr);            root->right = rightree;        }        else            root->right = NULL;        return root;    }}node* GetRightRoot(string str2)//建立右子树{    string leftStr,rightStr;    int position = -1;    node *leftree,*rightree,*root;    char ch;    int length = 0;    ch = porderStr[n];    n++;    position = str2.find(ch);    root = new node;    root->data = ch;    length = str2.length();    if(position == 0){        if(length > 1){            rightStr.assign(str2,1,length);            rightree = GetRightRoot(rightStr);            root->left = NULL;            root->right = rightree;            return root;        }        else{            root->left = NULL;            root->right = NULL;            return root;        }    }    else{        leftStr.assign(str2,0,position);        rightStr.assign(str2,position+1,length);        if(!leftStr.empty()){            leftree = GetLeftRoot(leftStr);            root->left = leftree;        }        else            root->left = NULL;        if(!rightStr.empty()){            rightree = GetRightRoot(rightStr);            root->right = rightree;        }        else            root->right = NULL;        return root;    }}void BuiltTree(node *root,int num)//用于还原二叉树{    int position = 0;    node *leftree,*rightree;    string leftStr,rightStr;    char ch = porderStr[n];    n++;    position = inorderStr.find(ch);    leftStr.assign(inorderStr,0,position);    rightStr.assign(inorderStr,position+1,num);    leftree = GetLeftRoot(leftStr);//调用递归函数，利用先序和中序序列特点    rightree = GetRightRoot(rightStr);    root->data = ch;    root->left = leftree;    root->right = rightree;}void postOrderBT(node *root){    if(root->left != NULL)        postOrderBT(root->left);    if(root->right != NULL)        postOrderBT(root->right);    cout<<root->data;}int main(){        string leftStr,rightStr;        int num;//中序序列的长度        node *root;        getline(cin,str);        while(!str.empty()){//判断是否到达输入的末端            root = new node;            n = 0;            num = str.length() / 2;//因为是整行输入所以需要分开先序和中序序列            porderStr.assign(str,0,num);            inorderStr.assign(str,num+1,2*num);            BuiltTree(root,num);//复原原二叉树            postOrderBT(root);//后序遍历二叉树；            cout<<endl;            getline(cin,str);        }        return 0;}

以上是菜菜的我自己按照自己的想法实现的，我知道网上可以找到很多现成的代码，但是我觉得重要的是学习的过程而不是结果，最重要的是自己学到的知识而不是仅仅就为了完成一道题，所以我向来在自己完成之前不会去看别人的代码，因为一旦看了别人的代码可能他的思想就会进入到自己的脑海，那么之后所做的也就只是将别人的想法实现这样根本就没有从本质上提高自己，但是在自己完成自后看看别人解决的方法倒是可以，或者说我觉得更是必须的，因为每一次的实践既是对自己检验也是给自己暴露不足取长补短的时候，所以在我们实现要做的后应该抱着学习的态度看看别人的做法，学习别人的长处这样才是对我们最大的帮助。

下面是我找到的别人实现的代码，很简洁，学习！

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>using namespace std;char pre[30];char in[30];void findPost(int root,int left,int right){    if(left>right)        return;    int i;    //left,right是针对于中缀表达式的    for(i = left;i<=right;i++)    {        if(in[i] == pre[root])        {            break;        }    }    //in[i]是根节点     findPost(root+1,left,i-1);    findPost(root + 1 + i - left,i+1,right);    printf("%c\n",in[i]);}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);#endif    while(scanf(" %s %s",pre,in)!=EOF)    {        findPost(0,0,strlen(in)-1);        printf("\n");    }}

根据树结构的特点，通过先序，后序，层序中的任意一个，加上中序就能实现重建二叉树。