PAT (Advanced Level) 1086. Tree Traversals Again (25) 根據先根周遊數組建立樹

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
      

注意到如果将pop看成數的空結點的話，輸入用例隻要在後面再加一個pop，該序列就與樹的先序周遊一樣。 我們可以用0表示空結點，則先根周遊序列轉換為1,2,3,0,0,4,0,0,5,6,0,0,0，利用該序列先根周遊建立樹即可。

/*2015.7.30cyq*/
#include <iostream>
#include <string>
#include <vector>
#include <fstream>
using namespace std;

//ifstream fin("case1.txt");
//#define cin fin

struct TNode{
	int val;
	TNode *left;
	TNode *right;
	TNode(int x):val(x),left(nullptr),right(nullptr){}//構造函數
};
//根據數組先根周遊建立樹
void createTree(TNode *&root,const vector<int> &ivec,int &cur){
	if(ivec[cur]==0){
		cur++;
	}else{//root指針指向一個new的結點，左右子結點先置為nullptr
		root=new TNode(ivec[cur]);
		cur++;
		createTree(root->left,ivec,cur);
		createTree(root->right,ivec,cur);
	}
}
//後序周遊
void postOrder(TNode *root,vector<int> &result){
	if(root!=nullptr){
		postOrder(root->left,result);
		postOrder(root->right,result);
		result.push_back(root->val);
	}
}
int main(){
	int N;
	cin>>N;
	vector<int> ivec;
	string s;
	int x;
	for(int i=0;i<2*N;i++){
		cin>>s;
		if(s=="Push"){
			cin>>x;
			ivec.push_back(x);
		}else
			ivec.push_back(0);//0表示null結點
	}
	ivec.push_back(0);//最後一個null結點

	TNode *root=nullptr;
	int cur=0;
	createTree(root,ivec,cur);

	vector<int> result;
	postOrder(root,result);

	cout<<result[0];
	for(auto it=result.begin()+1;it!=result.end();it++)
		cout<<" "<<*it;
	return 0;
}