1086. Tree Traversals Again (25)

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

#include <iostream>
#include <stack>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=50;
struct node
{
int data;
node* lchild;
node* rchild;
};
int pre[maxn],post[maxn],in[maxn];
int n;

node* create(int preL,int preR,int inL,int inR)
{
if(preL>preR)
{
return NULL;
}
node* root=new node;
root->data=pre[preL];
int k;
for(k=inL;k<=inR;k++)
{
if(in[k]==pre[preL])
{
break;
}
}
int numLeft=k-inL;
root->lchild=create(preL+1,preL+numLeft,inL,k-1);
root->rchild=create(preL+numLeft+1,preR,k+1,inR);
return root;
}
int num=0;
void postorder(node* root)
{
if(root==NULL)
{
return;
}
postorder(root->lchild);
postorder(root->rchild);
cout<<root->data;
num++;
if(num<n)
{
cout<<" ";
}
}
int main()
{
char str[5];
stack<int> st;
int x,preIndex=0,inIndex=0;
cin>>n;
for(int i=0;i<2*n;i++)
{
scanf("%s",str);
if(strcmp(str,"Push")==0)
{
cin>>x;
pre[preIndex++]=x;
st.push(x);
}
else
{
in[inIndex++]=st.top();
st.pop();
}
}
node* root=create(0,n-1,0,n-1);
postorder(root);
return 0;
}