PAT (Advanced Level) 1086. Tree Traversals Again (25) 根据先根遍历数组创建树

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看成数的空结点的话，输入用例只要在后面再加一个pop，该序列就与树的先序遍历一样。
我们可以用0表示空结点，则先根遍历序列转换为1,2,3,0,0,4,0,0,5,6,0,0,0，利用该序列先根遍历创建树即可。

/*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){}//构造函数
};
//根据数组先根遍历创建树
void createTree(TNode *&root,const vector<int> &ivec,int &cur){
if(ivec[cur]==0){
cur++;
}else{//root指针指向一个new的结点，左右子结点先置为nullptr
root=new TNode(ivec[cur]);
cur++;
createTree(root->left,ivec,cur);
createTree(root->right,ivec,cur);
}
}
//后序遍历
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结点
}
ivec.push_back(0);//最后一个null结点

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;
}