D 二叉树遍历（推荐)

二叉树遍历问题

Description

 

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

题目大意：

已知二叉树的先序和中序，求后序。

分析：

用递归对二叉树进行遍历。

代码：

#include <iostream>  
#include <cstring>  
using namespace std;  
  
char pre[27],mid[27];  
  
void T(int p1,int p2,int q1,int q2,int k )  //递归进行遍历
{  
    if (p1>p2)
        return;  
    for (k=q1;mid[k]!=pre[p1];++k);  
    T(p1+1,p1+k-q1,q1,k-1,0);  
    T(p1+k-q1+1,p2,k+1,q2,0);  
    cout<<mid[k];  
}  
  
int main()  
{  
    while(cin>>pre>>mid)
    {  
        int l=strlen(pre)-1;  
        T(0,l,0,l,0);  
        cout<<endl;  
    }  
    return 0;  
}  

心得：
二叉树问题看起来很容易，但要对先序、中序、以及后序有了解，才能更好的做二叉树的题目。二叉树问题无非就是两序找一序，一般都是用递归方法来实现，要好好学好递归，递归应用的方面有很多。