PAT 1086 Tree Traversals Again (25)

Posted 2020-11-20 mr-stn

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

 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 }