1086. Tree Traversals Again (25)

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


解题思路：由几个Push的操作可以组成树的先序遍历结果，而Pop的结果是树的中序遍历的结果，利用先序和中序即可构建一颗唯一的树，然后在后序遍历即可求解。

#include<iostream>
#include<vector>
#include<stack>
#include<cstdio>
#include<cstring>
using namespace std;
struct Tree{
	int data;
	Tree* left;
	Tree* right;
	Tree(){
		left=NULL;
		right=NULL;
	}
};
vector<int>preorder;
vector<int>inorder;
int n,cur;
int find(int key){
	for(int i=0;i<n;i++){
		if(inorder[i]==key){
			return i;
		}
	}
	return -1;
}
Tree* create(int start,int end){
	if(start>end)return NULL;
	Tree* t=new Tree();
	t->data=preorder[cur];
	cur++;
	int index=find(t->data);
	if(start!=end){
		t->left=create(start,index-1);
		t->right=create(index+1,end);
	}
	return t;
} 
int flag=0;
void postorder(Tree* t){
	if(!t)return;
	postorder(t->left);
	postorder(t->right);
	if(flag==0){
		printf("%d",t->data);
		flag=1;
	}else{
		printf(" %d",t->data);
	}
}
int main(){
	scanf("%d",&n);
	stack<int>s;
	int i,j;
	char str[5];
	int val;
	for(i=0;i<n*2;i++){
		scanf("%s",str);
		if(str[1]=='u'){
			scanf("%d",&val);
			s.push(val);
			preorder.push_back(val);
		}else{
			val=s.top();
			s.pop();
			inorder.push_back(val);
		}
	} 
	Tree* t=create(0,n-1);
	postorder(t);
	printf("\n");
	return 0;
}