[PAT] A1086 Tree Traversals Again
Posted yue36
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了[PAT] A1086 Tree Traversals Again相关的知识,希望对你有一定的参考价值。
题目大意
用栈的形式给出一棵二叉树的建立的顺序,求这棵二叉树的后序遍历
tips
string用printf输出:printf(“%s”, str.c_str());
AC代码
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include <cstdio>
#include <vector>
#include<algorithm>
#include<stack>
using namespace std;
vector<int>pre, in;
bool space = false;
void postorder(int root, int start, int end) {
if (start > end)return;
int i = start;
while (i < end && in[i] != pre[root])i++;
postorder(root + 1, start, i - 1);
postorder(root + (i - start) + 1, i + 1, end);
if (space == false) {
printf("%d", pre[root]);
space = true;
}
else printf(" %d", pre[root]);
}
int main() {
int i, n;
scanf("%d", &n);
stack<int>st;
string op;
int number;
for (i = 0;i < 2 * n;i++) {
cin >> op;
if (op == "Push") {
scanf("%d", &number);
st.push(number);
pre.push_back(number);
}
else {//op=="Pop"
in.push_back(st.top());
st.pop();
}
}
postorder(0, 0, n - 1);
return 0;
}
另
题目说结点编号1~N,保证了节点值互不相同。
如果在有多个节点的值相同的情况下,之前的代码会输出错误的结果,所以修改后的代码中添加了key作为索引,前中后序中均保存索引值,然后用value存储具体的值。(参考: https://blog.csdn.net/liuchuo/article/details/52181237 )代码如下:
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include <cstdio>
#include <vector>
#include <map>
#include<algorithm>
#include<stack>
#include <cstring>
using namespace std;
vector<int> pre, in, post, value;
void postorder(int root, int start, int end) {
if (start > end) return;
int i = start;
while (i < end && in[i] != pre[root]) i++;
postorder(root + 1, start, i - 1);
postorder(root + 1 + i - start, i + 1, end);
post.push_back(pre[root]);
}
int main() {
int n;
scanf("%d", &n);
char str[5];
stack<int> s;
int key = 0;
while (~scanf("%s", str)) {
if (strlen(str) == 4) {
int num;
scanf("%d", &num);
value.push_back(num);
pre.push_back(key);
s.push(key++);
} else {
in.push_back(s.top());
s.pop();
}
}
postorder(0, 0, n - 1);
printf("%d", value[post[0]]);
for (int i = 1; i < n; i++)
printf(" %d", value[post[i]]);
return 0;
}
以上是关于[PAT] A1086 Tree Traversals Again的主要内容,如果未能解决你的问题,请参考以下文章