uva 10701(简单二叉树)

最新推荐文章于 2019-12-31 16:15:05 发布

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最新推荐文章于 2019-12-31 16:15:05 发布

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分类专栏：树形结构-二叉树

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该问题要求根据给定的二叉树中序和前序遍历，计算出后序遍历的结果。题目提供了示例输入和输出，并指出每个测试用例包含树的节点数以及两个字符串，分别表示前序和中序遍历的顺序。

Problem F - Pre, in and post

Time Limit: 1 second

A common problem in data structures is to determine the traversal of a binary tree. There are three classic ways to do it:

Pre-order: You must visit in sequence the root, left subtree and the right subtree.
In-order: You must visit in sequence the left subtree, the root and the right subtree.
Post-order: You must visit in sequence the left subtree, the right subtree and the root.

See the picture below:

The pre, in and post-order traversal are, respectively, ABCDEF, CBAEDF and CBEFDA. In this problem, you must compute the post-order traversal of a binary tree given its in-order and pre-order traversals.

Input

The input set consists of a positive number C ≤ 2000, that gives the number of test cases and C lines, one for each test case. Each test case starts with a number 1 ≤ N ≤ 52, the number of nodes in this binary tree. After, there will be two strings S₁ and S₂ that describe the pre-order and in-order traversal of the tree. The nodes of the tree are labeled with different characters in the range a..z and A..Z. The values of N, S₁ and S₂ are separeted by a blank space.

Output

For each input set, you should output a line containing the post-order transversal for the current tree.

Sample input

3
3 xYz Yxz
3 abc cba
6 ABCDEF CBAEDF

Sample output

Yzx
cba
CBEFDA

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
using namespace std;

const int maxn = 60;
struct Node{
	char c;
	int son[2];
}node[maxn];
int N , cnt;
string s1 , s2;

void initial(){
	for(int i = 0;i < maxn;i++){
		node[i].c = '0';
		node[i].son[0] = -1;
		node[i].son[1] = -1;
	}
	s1.clear();
	s2.clear();
	cnt = 0;
}

void readcase(){
	cin >> N >> s1 >> s2;
}

int get_tree(string str2){
	if(s1.length() > 0){
		int n = cnt++;
		node[n].c = s1[0];
		s1 = s1.substr(1 , s1.length()-1);
		for(int i = 0;i < str2.length();i++){
			if(node[n].c == str2[i]){
				if(i > 0){
					node[n].son[0] = get_tree(str2.substr(0 , i));
				}
				if(str2.length()-i-1 > 0){
					node[n].son[1] = get_tree(str2.substr(i+1 , str2.length()-i-1));
				}
				return n;
			}
		}
		return -1;
	}
}

void print(int i){
	if(node[i].son[0] != -1){
		print(node[i].son[0]);
	}
	if(node[i].son[1] != -1){
		print(node[i].son[1]);
	}
	printf("%c" , node[i].c);
}

void computing(){
	int head = get_tree(s2);
	print(0);
	printf("\n");
}

int main(){
	int C;
	cin >> C;
	while(C--){
		initial();
		readcase();
		computing();
	}
	return 0;
}