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
1 #include<iostream> 2 #include<vector> 3 #include<stack> 4 using namespace std; 5 vector<int> in, post, pre; 6 void getPost(int prel, int prer, int inl, int inr){ 7 if(prel>prer) return; 8 int i=inl; 9 while(i<=inr && in[i]!=pre[prel]) i++; 10 getPost(prel+1, prel+i-inl, inl, i-1); 11 getPost(prel+i-inl+1, prer, i+1, inr); 12 post.push_back(pre[prel]); 13 14 } 15 int main(){ 16 int n, i; 17 cin>>n; 18 stack<int> s; 19 for(i=0; i<2*n; i++){ 20 char op[5]; 21 int t; 22 scanf("%s", op); 23 if(op[1]=='u'){ 24 scanf("%d", &t); 25 pre.push_back(t); 26 s.push(t); 27 }else{ 28 in.push_back(s.top()); 29 s.pop(); 30 } 31 } 32 getPost(0, n-1, 0, n-1); 33 cout<<post[0]; 34 for(i=1; i<n; i++) cout<<" "<<post[i]; 35 return 0; 36 }