PAT 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.

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

 

压栈顺序是前序遍历的顺序， 出栈可以得到中序遍历顺序； 所以通过前序和中序遍历得到后序遍历

思路：也是找根节点和子树节点个数的问题， 和https://i.cnblogs.com/EditPosts.aspx?postid=9214298 类似

#include<iostream>
#include<vector>
#include<stack>
using namespace std;
vector<int> in, post, pre;
void getPost(int prel, int prer, int inl, int inr){
    if(prel>prer) return;
    int i=inl;
    while(i<=inr && in[i]!=pre[prel]) i++;
    getPost(prel+1, prel+i-inl, inl, i-1);
    getPost(prel+i-inl+1, prer, i+1, inr);
    post.push_back(pre[prel]);
 
}
int main(){
  int n, i;
  scanf("%d", &n);
  stack<int> s;
  for(i=0; i<2*n; i++){
    char op[5];
    int t;
    scanf("%s", op);
    if(op[1]=='u'){
      scanf("%d", &t);
      pre.push_back(t);
      s.push(t);
    }else{
      in.push_back(s.top());
      s.pop();
    }
  }
  getPost(0, n-1, 0, n-1);
  printf("%d", post[0]);
  for(i=1; i<n; i++) printf(" %d", post[i]);
  return 0;
}