一 原题
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
三 代码
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.
/*
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();
}