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

分析：
题目通过栈的方式给出二叉树的遍历过程，Push 就是先序遍历，Pop就是中序遍历。
可以开二个数组
作为保存两个序列 pre[maxn] ino[maxn]
然后根据两个序列来构造二叉树（递归的方式）
注意终止条件： if(preL>preR) return NULL;
以及各个边界

#include<bits/stdc++.h>
using namespace std;
const int maxn=50;
int pre[maxn],ino[maxn],post[maxn];
struct node
{
  int data;
  node *lchild;
  node *rchild;
};
node* create(int preL,int preR,int inoL,int inoR)
{
    if(preL>preR) return NULL;
    node *root= new node;
    root->data=pre[preL];
    int k;
    for(k=inoL;k<=inoR;k++)
    {
        if(ino[k]==pre[preL])
            break;
    }
    int numleft=k-inoL;
    root->lchild=create(preL+1,preL+numleft,inoL,k-1);
    root->rchild=create(preL+numleft+1,preR,k+1,inoR);
    return root;
}
int num=0,n;
void postorder(node *root)
{
     if(root==NULL) return;
     postorder(root->lchild);
     postorder(root->rchild);
     printf("%d",root->data);
     num++;
     if(num<n) printf(" ");
};
int main()
{
     scanf("%d",&n);
     int val,index_p=0,index_i=0;
     char str[5];
     stack<int> s;
     for(int i=0;i<2*n;i++)
     {
         scanf("%s",str);
         if(strcmp(str,"Push")==0)
         {
             scanf("%d",&val);
             pre[index_p++]=val;
             s.push(val);
         }
         else
         {
             ino[index_i++]=s.top();
             s.pop();
         }
     }
     node *root=create(0,n-1,0,n-1);
     postorder(root);
     return 0;
}