A1086

本題的關鍵是對所給樣例的理解，看了算法筆記才明白，push的次序是先序遍歷順序，pop的次序是中序遍歷順序.其餘跟A1020一樣.
A1020通過後序、中序遍歷獲取二叉樹
A1086通過先序、中序遍歷獲取二叉樹

#include<cstdio>
#include<cstdlib>
#include<string.h>
#include<math.h>
#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<algorithm>
using namespace std;
const int maxn=35;
int flag,n,preorder[maxn],inorder[maxn],pre=1,in=1;
struct node{
	int data;
	node* lchild;
	node* rchild;
};
node* binary(int preL,int preR,int inL,int inR){
	if(preL>preR){
		return NULL;
	}
	node* root=new node;
	root->data=preorder[preL];
	int k=inL;
	while(inorder[k]!=preorder[preL]){
		k++;
	}
	int left_len=k-inL;
	int right_len=inR-k;
	root->lchild=binary(preL+1,preL+left_len,inL,k-1);
	root->rchild=binary(preR-right_len+1,preR,k+1,inR);
	return root;                                                      //忘了將根結點返回出去 
}
void dfs(node* root){
	if(root==NULL){
		return;
	}
	dfs(root->lchild);
	dfs(root->rchild);
	if(root->data==flag){
		printf("%d",root->data);
	}else{
		printf("%d ",root->data);
	}
}
int main(){
	#ifdef ONLINE_JUDGE
	#else
		freopen("1.txt","r",stdin);
	#endif
	char str[5];
	stack<int> s;
	scanf("%d",&n);
	for(int i=0;i<n*2;i++){
		scanf("%s",str);
		if(strcmp(str,"Push")==0){
			scanf("%d",&preorder[pre++]);
			s.push(preorder[pre-1]);
		}else{
			inorder[in++]=s.top();
			s.pop();
		}
	}
	node* root=binary(1,pre-1,1,in-1);
	flag=root->data;
	dfs(root);
	return 0;
}