PAT 1086. Tree Traversals Again

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

分析

这道题是由树的前序遍历和中序遍历来获得树的后序遍历的意思，首先由输入的数据的按输入顺序储存到数组中获得先序遍历的顺序，然后按输入的操作，即每次遇到pop就pop出的元素储存在一个数组里获得中序遍历的结果。接下来就是由这两个部分获得后序遍历了，首先先序遍历的第一个元素一定是该子树的根节点，然后在中序遍历的数组中找到该数，由中序遍历的性质可知，该数左边的是根节点的左子树元素，右边的是右子树的元素，接下来就是对左右两边递归调用前面的部分了，不过要先调用右子树部分，因为最后输出是倒着输出的。

#include<iostream>
#include<vector>
#include<stack>
#include<algorithm>
using namespace std;
vector<int> preorder,inorder,postorder;
void getpost(int root,int s,int e){
	if(s>e) return ;
	postorder.push_back(preorder[root]);
	int index=find(inorder.begin(),inorder.end(),preorder[root])-inorder.begin();
	getpost(root+index-s+1,index+1,e);
	getpost(root+1,s,index-1);
}
int main(){
	int n,t;
	string order;
	cin>>n;
	stack<int> s;
	for(int i=0;i<2*n;i++){
		cin>>order;
		if(order=="Push"){
		   cin>>t;
		   preorder.push_back(t);
		   s.push(t);
		}else{
			int u=s.top();
			s.pop();
			inorder.push_back(u);
		}
	}
	getpost(0,0,n-1);
	for(int i=n-1;i>=0;i--)
	    i==n-1?cout<<postorder[i]:cout<<" "<<postorder[i];
	return 0;
}