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 思路：本题出现意外，index不能再外面定义，在PAT OJ中定义会出现错误，另外本道题关于数学的知识有两点内容：1：有n个顶点，如果确定唯一一棵树的话进出栈的顺序次数是2*n2: 另外进栈的顺序是先序，出栈的顺序是中序。 这两点规律需要记住。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <stack>
5 using namespace std;
6 //定义静态链表
7 #define MAX 50
8 struct node
9 {
10 int data;
11 node* lchild;
12 node* rchild;
13 };
14 int pre[MAX];
15 int in[MAX];
16 int post[MAX];
17 int ind=0;
18
19 node * Create(int preL,int preR,int inL,int inR)
20 {
21 if(preL>preR)
22 return NULL;
23 node * root=new node;
24 root->data=pre[preL];
25 int k;
26 for(k=inL;k<=inR;k++)
27 {
28 if(in[k]==root->data)
29 {
30 break;
31 }
32 }
33 int numleft=k-inL;
34 root->lchild=Create(preL+1,preL+numleft,inL,k-1);
35 root->rchild=Create(preL+numleft+1,preR,k+1,inR);
36 return root;
37 }
38
39 void Post(node * root)
40 {
41 if(root==NULL)
42 return ;
43 Post(root->lchild);
44 Post(root->rchild);
45 post[ind++]=root->data;
46 }
47
48
49
50 int main(int argc, char *argv[])
51 {
52 node * root;
53 int n;
54 scanf("%d",&n);
55 stack<int>st;
56 char str[10];
57 int prept=0,inpt=0;
58 for(int i=0;i<2*n;i++)
59 {
60 scanf("%s",str);
61 if(!strcmp(str,"Push"))
62 {
63 int number;
64 scanf("%d",&number);
65 st.push(number);
66 pre[prept++]=number;
67 }
68 else
69 {
70 int number=st.top();
71 st.pop();
72 in[inpt++]=number;
73 }
74 }
75 root=Create(0,n-1,0,n-1);
76 Post(root);
77 /*
78 for(int i=0;i<n;i++)
79 cout<<pre[i]<<" ";
80 cout<<endl;
81 for(int i=0;i<n;i++)
82 cout<<in[i]<<" ";
83 cout<<endl;*/
84 for(int i=0;i<n;i++)
85 {
86 printf("%d",post[i]);
87 if(i<n-1)
88 putchar(‘ ‘);
89 }
90 putchar(‘\n‘);
91 return 0;
92 }