PAT甲级1086

1086. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

#include<cstdio>
#include<iostream>
#include<string>
#include<stack>
using namespace std; 
struct node
{
  int data;
  node*lchild, *rchild;
  node():lchild(NULL),rchild(NULL){}
}*root;
bool flag = false;
void postOrder(node*root)
{
  if (!root)
  {
    return;
  }
  postOrder(root->lchild);
  postOrder(root->rchild);
  if (!flag)
  {
    printf("%d", root->data);
    flag = true;
  }
  else
    printf(" %d", root->data);
}
int main()
{
  int N;
  scanf("%d", &N);
  int i = 0;
  string s; int t;
  stack<node*> st;
  bool tag = false;
  node *ROOT;
  while (i < N)
  {
    cin >> s;
    if (s == "Push")
    {
      cin >> t;
      node* newNode = new node;
      newNode->data = t;
      if (!tag)
      {
        if (!root)
        {
          root = newNode;
          ROOT = root;
        }
        else
        {
          root->lchild = newNode;
          root = root->lchild;
        }
      }
      else
      {
        root->rchild = newNode;
        root = root->rchild;
        tag = false;
      }
      st.push(newNode);
      i++;
    }
    else
    {
      tag = true;
      root = st.top();
      st.pop();
    }
  }
  postOrder(ROOT);
  return 0;
}