PAT_A1086#Tree Traversals Again

Source:

PAT A1086 Tree Traversals Again (25 分)

Description:

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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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

Keys:

• 二叉树的建立和遍历

Code:

 1 /*
 2 time: 2019-06-30 14:34:48
 3 problem: PAT_A1086#Tree Traversals Again
 4 AC: 24:16
 5 
 6 题目大意：
 7 给出中序遍历的出栈和入栈操作，打印后序遍历
 8 
 9 基本思路：
10 模拟二叉树的遍历过程，Push就是存在子树，Pop就是空子树
11 */
12 #include<cstdio>
13 #include<string>
14 #include<iostream>
15 using namespace std;
16 int n;
17 
18 void InOrder()
19 {
20     string op;
21     int data;
22     cin >> op;
23     if(op == "Push")
24         scanf("%d", &data);
25     else
26         return;
27     static int pt=0;
28     InOrder();
29     InOrder();
30     printf("%d%c", data, ++pt==n?'\n':' ');
31 }
32 
33 int main()
34 {
35 #ifdef ONLINE_JUDGE
36 #else
37     freopen("Test.txt", "r", stdin);
38 #endif // ONLINE_JUDGE
39 
40     scanf("%d", &n);
41     InOrder();
42 
43     return 0;
44 }