[PAT] A1086 Tree Traversals Again

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; 
}