1086 Tree Traversals Again——PAT甲级

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1

 solution：
 

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
typedef long long ll;
int n;
typedef struct node *tree;
struct node
{
	int val;
	tree left,right;
};
const int maxn=1e5+5;
int pre[maxn],in[maxn];
int cnt1=0,cnt2=0;
vector<int>ans;
tree build(int pre[],int in[],int n)
{
	if(!n)return NULL;
	tree root=new node;
	root->val=pre[0];
	root->left=root->right=NULL;
	int i;
	for(i=0;i<n;i++)
	{
		if(pre[0]==in[i])break;
	}
	root->left=build(pre+1,in,i);
	root->right=build(pre+1+i,in+1+i,n-i-1);
	return root;
}
void postorder(tree t)
{
	if(t==NULL)return;
	postorder(t->left);
	postorder(t->right);
	ans.push_back(t->val);
}
void output()
{
	//cout<<ans.size()<<endl;
	for(int i=0;i<ans.size();i++)
	{
		if(i)cout<<' ';
		cout<<ans[i];
	}
}
int main()
{
	cin>>n;
	stack<int>st;
	for(int i=1;i<=2*n;i++)
	{
		string s;cin>>s;
		if(s=="Push")
		{
			int t;cin>>t;
			pre[cnt1++]=t;
			st.push(t);
		}
		else
		{
			int tmp=st.top();st.pop();
			in[cnt2++]=tmp;
		}
	}
	
	//for(int i=0;i<n;i++)cout<<pre[i]<<' ';cout<<endl;
	//for(int i=0;i<n;i++)cout<<in[i]<<' ';cout<<endl;
	
	tree root=new node;
	root=build(pre,in,n);
	postorder(root);
	output();
}

 

单纯的依靠前序和中序建后序