usaco3.4.1 American Heritage

一 原题

American Heritage

Farmer John takes the heritage of his cows very seriously. He is not, however, a truly fine bookkeeper. He keeps his cow genealogies as binary trees and, instead of writing them in graphic form, he records
them in the more linear `tree in-order' and `tree pre-order' notations.
Your job is to create the `tree post-order' notation of a cow's heritage after being given the in-order and pre-order notations. Each cow name is encoded as a unique letter. (You may already know that you
can frequently reconstruct a tree from any two of the ordered traversals.) Obviously, the trees will have no more than 26 nodes.
Here is a graphical representation of the tree used in the sample input and output:

C
/   \
/     \
B       G
/ \     /
A   D   H
/ \
E   F

The in-order traversal of this tree prints the left sub-tree, the root, and the right sub-tree.
The pre-order traversal of this tree prints the root, the left sub-tree, and the right sub-tree.
The post-order traversal of this tree print the left sub-tree, the right sub-tree, and the root.

PROGRAM NAME: heritage

INPUT FORMAT

Line 1:	The in-order representation of a tree.
Line 2:	The pre-order representation of that same tree.

SAMPLE INPUT (file heritage.in)

ABEDFCHG
CBADEFGH

OUTPUT FORMAT

A single line with the post-order representation of the tree.

SAMPLE OUTPUT (file heritage.out)

AEFDBHGC

二 分析

给定二叉树的前序遍历和中序遍历，求后序遍历。

三 代码

运行结果：

USER: Qi Shen [maxkibb3]
TASK: heritage
LANG: C++

Compiling...
Compile: OK

Executing...
Test 1: TEST OK [0.000 secs, 4188 KB]
Test 2: TEST OK [0.000 secs, 4188 KB]
Test 3: TEST OK [0.000 secs, 4188 KB]
Test 4: TEST OK [0.000 secs, 4188 KB]
Test 5: TEST OK [0.000 secs, 4188 KB]
Test 6: TEST OK [0.000 secs, 4188 KB]
Test 7: TEST OK [0.000 secs, 4188 KB]
Test 8: TEST OK [0.000 secs, 4188 KB]
Test 9: TEST OK [0.000 secs, 4188 KB]

All tests OK.

YOUR PROGRAM ('heritage') WORKED FIRST TIME! That's fantastic
-- and a rare thing. Please accept these special automated
congratulations.

AC代码：

/*
ID:maxkibb3
LANG:C++
PROB:heritage
*/

#include<iostream>
#include<fstream>
using namespace std;

string in, pre, post;
ifstream fin;
ofstream fout;
int pre_idx;

struct Node {
char val;
Node *left, *right;
};

Node *build_tree(int l, int r) {
if(l > r) return NULL;
Node *root = new Node();
pre_idx++;
int in_idx = in.find(pre[pre_idx]);
root->val = pre[pre_idx];
if(l != r) {
root->left = build_tree(l, in_idx - 1);
root->right = build_tree(in_idx + 1, r);
}
return root;
}

void post_dfs(Node *node) {
if(!node) return;
post_dfs(node->left);
post_dfs(node->right);
fout << node->val;
}

int main() {
fin.open("heritage.in");
fout.open("heritage.out");

fin >> in >> pre;

pre_idx = -1;
Node *root = build_tree(0, in.length() - 1);
post_dfs(root);
fout << endl;
delete root;

fin.close();
fout.close();
}