PAT甲级——A1086 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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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

 //这道题就是已知前序和中序遍历，得到后序遍历
 //push为前序遍历，pop为中序遍历
 #include <iostream>
 #include <vector>
 #include <stack>
 #include <string>
 using namespace std;
 int N;
 vector<int>preOrder, inOrder, posOrder;
 struct Node
 {
     int val;
     Node *l, *r;
     Node(int a = ) :val(a), l(nullptr), r(nullptr) {};
 };
 Node* createTree(int preL, int preR, int inL, int inR)
 {
     if (preL > preR)
         return nullptr;
     Node* root = new Node(preOrder[preL]);
     int i;
     for (i = inL; i <= inR; ++i)//找到根节点
         if (inOrder[i] == preOrder[preL])
             break;
     int num = i - inL;
     root->l = createTree(preL + , preL + num, inL, i - );
     root->r = createTree(preL + num + , preR, i + , inR);
     return root;
 }
 void posOrderTree(Node *root)
 {
     if (root == nullptr)
         return;
     posOrderTree(root->l);
     posOrderTree(root->r);
     posOrder.push_back(root->val);
 }
 int main()
 {

     cin >> N;
     string str;
     stack<int>s;
     int a;
     for (int i = ; i < *N; ++i)
     {
         cin >> str;
         if (str == "Push")
         {
             cin >> a;
             s.push(a);
             preOrder.push_back(a);
         }
         else
         {
             inOrder.push_back(s.top());
             s.pop();
         }
     }
     Node* root = createTree(, N - , , N - );
     posOrderTree(root);
     for (int i = ; i < N; ++i)
         cout << posOrder[i] << (i == N -  ? "" : " ");
     return ;
 }