UVA - 536 Tree Recovery
来源:互联网 发布:网络推广访问者 编辑:程序博客网 时间:2024/04/30 13:49
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 ABCDEFGBCAD CBAD
Sample Output
ACBFGEDCDAB
二叉树+BFS题目,略水
#include <iostream>#include <cstring>#include <vector>#include <string>#include <algorithm>#define MAXN 256using namespace std;string PreOrder, InOrder;vector<char> PostOrder;char Lch[MAXN], Rch[MAXN];int Build_tree(int L1, int R1, int L2, int R2){if (L2 > R2)return false;char root = PreOrder[L1];int p = L2;while (root != InOrder[p])p++;int cnt = p - L2;Lch[root] = Build_tree(L1 + 1, L1 + cnt, L2, L2 + cnt - 1);Rch[root] = Build_tree(L1 + cnt + 1, R1, L2 + cnt + 1, R2);return root;}void dfs(int n){if (Lch[n]) dfs(Lch[n]);if (Rch[n]) dfs(Rch[n]);PostOrder.push_back(n);}int main(){while (cin >> PreOrder >> InOrder){memset(Lch, 0, sizeof(Lch));memset(Rch, 0, sizeof(Rch));char root = Build_tree(0, PreOrder.length() - 1, 0, InOrder.length() - 1);PostOrder.clear();dfs(root);for (size_t i = 0; i < PostOrder.size(); i++)cout << PostOrder[i];cout << endl;}return 0;}
- UVA 536 - Tree Recovery
- uva 536Tree Recovery
- uva 536 - Tree Recovery
- UVA 536 Tree Recovery
- UVA 536 Tree Recovery
- UVa, 536 Tree Recovery
- uva 536 - Tree Recovery
- UVa 536 - Tree Recovery
- UVA - 536 Tree Recovery
- UVa 536 Tree Recovery
- UVa 536 - Tree Recovery
- UVA - 536 Tree Recovery
- Uva - 536 - Tree Recovery
- Tree Recovery-UVA 536
- uva 536 - Tree Recovery
- Uva-536 Tree Recovery
- UVA - 536 Tree Recovery
- UVA 536 Tree Recovery
- HDU 2021
- 获取手机的储存空间
- HDU 2022
- 一致性。
- HDU 2023
- UVA - 536 Tree Recovery
- HDU 2024
- 程立支付宝首席架构师,视频演讲地址,
- HDU 2025
- 关于ViewPager的懒加载问题
- Javascript的奇怪用法
- 我是一名程序员
- 区分struct和typedef struct
- http、TCP/IP协议与socket之间的区别