zoj 1944 Tree Recovery (二叉樹）

Tree Recovery

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Time Limit: 2 Seconds Memory Limit: 65536 KB

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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

這道題和之前做的重建二叉樹 思路基本一致，那道題是已知中序和後序遍歷，求前序遍歷，這道題就是已知前序遍歷和中序遍歷，求後序遍歷，由前序遍歷可以得到這棵樹的根節點，由中序遍歷把這棵樹分成兩部分，然後再進行遞歸求解；

下面是代碼：

#include 
#include 
#include 
using namespace std;
struct node
{
    char value;
    node *lchild,*rchild;//左右子樹
};
node *newnode(char c)
{
    node *p;
    p=(node *)malloc(sizeof(node));
    p->value=c;
    p->lchild=p->rchild=NULL;
    return p;
}
node *rebulid(char *pre,char *in,int n)
{
    if(n==0) return NULL;
    int l_len,r_len,i;
    char s=pre[0];
    node *root=newnode(s);
    for(i=0;i0) root->lchild=rebulid(pre+1,in,l_len);//遞歸求解
    if(r_len>0) root->rchild=rebulid(pre+l_len+1,in+l_len+1,r_len);
    return root;
}
void postorder(node *p)
{
    if(p==NULL) return;
    postorder(p->lchild);
    postorder(p->rchild);
    printf("%c",p->value);
}
int main()
{
    char preorder[30],inorder[30];
    while(scanf("%s%s",preorder,inorder)!=EOF)
    {
        node *root=rebulid(preorder,inorder,strlen(preorder));
        postorder(root);
        printf("\n");
    }
    return 0;
}