PAT 数据结构 03-树3. 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:

6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop

Sample Output:

3 4 2 6 5 1
/*2015.7.8*///03-3 先根遍历创建树，使用指针的引用和迭代器的引用，递归#include <iostream>#include <math.h>#include <vector>#include <queue>#include <string>using namespace std;struct TNode{int val;TNode* left;TNode* right;TNode(int x):val(x),left(nullptr),right(nullptr){}};void createTree(TNode *&root,vector<int>::iterator &it){if((*it)==0){root=nullptr;it++;}else{//数据不为0，root指针指向一个new的结点root=new TNode(*it);it++;createTree(root->left,it);createTree(root->right,it);}}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;N=2*N;string s;int x;vector<int> seq(N+1);for(int i=0;i<N;i++){cin>>s;if(s=="Push"){cin>>x;seq[i]=x;}elseseq[i]=0; //0表示NULL}seq[N]=0;vector<int>::iterator it=seq.begin();TNode* root;createTree(root,it);vector<int> result;postOrder(root,result);int n=result.size();cout<<result[0];for(int i=1;i<n;i++)cout<<" "<<result[i];return 0;}