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
1 #include <stdio.h>
2 #include <stdlib.h>
3 #include <iostream>
4 #include <string.h>
5 #include <math.h>
6 #include <algorithm>
7 #include <string>
8 #include <stack>
9 #include <queue>
10 using namespace std;
11 const int maxn=33;
12 int n;
13 int pre[maxn],in[maxn];
14 struct node{
15 int data;
16 node *left;
17 node *right;
18 };
19 //给一个先序和中序，求层次遍历
20 node *create(int a,int b,int c,int d)
21 {
22 if(a>b)return NULL;
23 node *temp=new node;
24 temp->data=pre[a];
25 int k=0;
26 for(;k+c<=d;k++)
27 {
28 if(in[c+k]==pre[a])break;
29 }
30 temp->left=create(a+1,a+k,c,c+k-1);
31 temp->right=create(a+k+1,b,c+k+1,d);
32 return temp;
33 }
34 //后序
35 int printcount=0;
36 void postorder(node * root)
37 {
38 if(root==NULL)return;
39 postorder(root->left);
40 postorder(root->right);
41 printf("%d",root->data);
42 printcount++;
43 if(printcount<n)printf(" ");
44 }
45 void BFS(node * root)
46 {
47 int count=0;
48 queue<node*> q;
49 q.push(root);
50 while(!q.empty())
51 {
52 node* now=q.front();
53 q.pop();
54 printf("%d",now->data);
55 count++;
56 if(count<n)printf(" ");
57 if(now->left!=NULL)q.push(now->left);
58 if(now->right!=NULL)q.push(now->right);
59 }
60 }
61 int main(){
62 int temp,num=0,innum=0;
63 stack<int> st;
64 scanf("%d",&n);
65 for(int i=0;i<n*2;i++)
66 {
67 char a[5];
68 scanf("%s",a);
69 if(strcmp(a,"Push")==0)
70 {
71 scanf("%d",&temp);
72 pre[num++]=temp;
73 st.push(temp);
74 }//pop
75 else
76 {
77 in[innum++]=st.top();
78 st.pop();
79 }
80 }
81 //根据前序中序 建树
82
83 node* root=create(0,n-1,0,n-1);
84 postorder(root);
85 return 0;
86 }