03-树3 Tree Traversals Again(25 分)
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
题意:通过栈和中序遍历的方式输入一棵树,然后需要输出他的后序遍历
AC代码:
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <list>
#include <map>
#include <stack>
#include <queue>
using namespace std;
#define ll long long
typedef struct Tree
{
int data;
Tree *l,*r;
}*BiTree;
stack<int> q;
int sum;
int n;
void Creat(BiTree &T)
{
if(sum == 2*n)
return;
char x[11];
cin >> x;
sum++;
if(x[1] == 'u')
{
int x;
cin >> x;
q.push(x);
T = new Tree;
T->l = T->r = NULL;
Creat(T->l);
T->data = q.top();
q.pop();
Creat(T->r);
}
return;
}
void endorder(BiTree T)
{
if(!T)
return;
endorder(T->l);
endorder(T->r);
if(sum == 2*n)
cout << T->data;
else
cout <<' '<< T->data;
sum++;
}
int main()
{
BiTree T;
cin >> n;
sum = 0;
Creat(T);
endorder(T);
cout << endl;
//cout << "AC" <<endl;
return 0;
}
反思:一开始把栈写成了队列......