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.

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

3 4 2 6 5 1

思路
在1020的基礎上稍微增加了難度。
1.仔細觀察發現這道題中棧的壓入操作其實是樹的前序遍歷，彈出操作是樹的中序遍歷
2.那麽可以根據輸入操作的不同構造出樹的前序序列和中序序列，由此轉化為類似pat1020 的問題了。
代碼

#include<iostream>
#include<vector>
using namespace std;
vector<int> preorder(31);
vector<int> inorder(31);
vector<int> mypostorder(31);
int index = 1;

void postorder(int pfirst,int plast,int ifirst,int ilast )
{
      if(pfirst > plast || ifirst > ilast)
        return;
      int i = 0;
      while(preorder[pfirst] != inorder[ifirst + i]) i++; //find root‘s index in inorder sequence

      postorder(pfirst + 1,pfirst + i,ifirst,ifirst + i);
      postorder(pfirst + i + 1,plast,ifirst + i + 1,ilast);
      mypostorder[index++] = preorder[pfirst];
}

int main()
{
  int N;
  while(cin >> N)
  {
      vector<int> stk;
      int in = 1,out = 1;
      //input
      while(in <= N || out <= N)
      {
        string command;
        int value;
        cin >> command ;
        if(command[1] == ‘u‘)
        {
            cin >> value;
            preorder[in++] = value;
            stk.push_back(value);
        }
        else
        {
            inorder[out++] = stk.back();
            stk.pop_back();
        }
      }

      //handle
      postorder(1,N ,1,N);
      //output
      int j = 1;
      cout << mypostorder[j];
      for(int j = 2;j <= N;j++)
        cout <<" " << mypostorder[j];
      cout << endl;
  }
}

PAT1086:Tree Traversals Again