POJ 2255 Tree Recovery

Tree Recovery

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9879		Accepted: 6210

Description

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 ABCDEFG
BCAD CBAD

Sample Output

ACBFGED
CDAB
题目大意：给出二叉树的先序，中序序列，求二叉树的后序序列。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stack>
using namespace std;

typedef struct node
{
    char data;
    node *lchild;
    node *rchild;
    node()
    {
        lchild = rchild = NULL;
    }
}TreeNode;

//二叉树的非递归后序遍历
TreeNode *BuildTree(char *pre, char *in, int n)
{
    TreeNode *pNode;
    int k;
    char *p;
    if (n <= 0)
    {
        return NULL;
    }
    pNode = new TreeNode;//新建节点
    pNode->data = *pre;
    for (p = in; p < in + n; p++)//在中序序列中找等于*pre的位置k
    {
        if (*p == *pre)//pre指向根节点
        {
            break;//找到后退出，从而确定根节点在in中的位置
        }
    }
    k = p - in;//左子树的长度
    pNode->lchild = BuildTree(pre + 1, in, k);//递归构造左子树
    pNode->rchild = BuildTree(pre + k + 1, p + 1, n - k - 1);//递归构造右子树
    return pNode;//返回根节点
}

//二叉树的非递归后序遍历
void PostOrder(TreeNode *pRoot)
{
    stack<TreeNode *> Stack;
    TreeNode *p = pRoot;
    TreeNode *q;
    do 
    {
        while(p != NULL)
        {
            Stack.push(p);
            p = p->lchild;
        }
        q = NULL;
        while(!Stack.empty())
        {
            p = Stack.top();
            if (q == Stack.top()->rchild)
            {
                printf("%c", Stack.top()->data);
                q = Stack.top();
                Stack.pop();
            }
            else
            {
                p = p->rchild;
                break;
            }
        }
    } while (!Stack.empty());
}

//删除二叉树节点
void DeleteNode(TreeNode *pRoot)
{
    if (pRoot != NULL)
    {
        DeleteNode(pRoot->lchild);
        DeleteNode(pRoot->rchild);
    }
    delete pRoot;
}

int main()
{
    char pre[30], in[30];
    while(scanf("%s%s", pre, in) != EOF)
    {
        TreeNode *pRoot = BuildTree(pre, in, strlen(in));
        PostOrder(pRoot);
        printf("
");
        DeleteNode(pRoot);
    }
    return 0;
}