由先序和中序求后序

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

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
题意：按先序输入，pop出的顺序是中序，输出后序。

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<stack>
 6 using namespace std;
 7 struct TNode
 8 {
 9     int left;
10     int right;
11 } Tree[35];//树的节点最多不超过 30
12 int s;//用来保存树的根的值  因为好像第一个测试用例的树根不是  1
13 void Create()
14 {  memset(Tree,0,sizeof(Tree));
15     int n;
16     scanf("%d",&n);
17     cin.get();
18     stack<int >S;
19     int top;
20     char str[10];
21     int data;
22     scanf("%s %d\n",str,&data);
23     S.push(data);
24     s=top=S.top();
25   
26     for (int i=1; i<2*n; i++)//树根已入栈，故从i=1开始
27     {
28         scanf("%s",str);
29         if(strlen(str)!=3)//注意这个if结构
30         {
31             scanf("%d",&data);
32             S.push(data);
33             if(Tree[top].left==0)//左子树空，将data放入左子树
34                 Tree[top].left=data;
35             else //左子树已被使用
36                 Tree[top].right=data;
37             top=S.top();//更新top
38         }
39         else
40         {//注意这个大括号里的代码
41             top=S.top();
42             S.pop();
43         }
44     }
45 }
46 int k=0;
47 void Merge(  int n)//控制输入输出格式
48 {
49 
50     if(Tree[n].left!=0)
51         Merge(Tree[n].left);
52     if(Tree[n].right!=0)
53         Merge(Tree[n].right);
54     if(k==0)
55     {
56         printf("%d",n);
57         k=1;
58     }
59     else
60         printf(" %d",n);
61 }
62 int main()
63 {
64     Create();
65     Merge(s);
66     return 0;
67 
}