Source:
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:
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 }
原文地址：https://www.cnblogs.com/blue-lin/p/11109674.html