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 $N$ ( $\leq 30$ ) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to $N$ ). Then $2 N$ 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

题目的意思就是输入一颗二叉树的先序遍历和中西医遍历，让你输出后序遍历

Push操作的是先序遍历 Pop操作的是中序遍历

我们可以用一个堆栈来模拟输入的Push和Pop操作。其中用到了c++中的strcmp函数

注意：strcmp函数只对char类型进行比较所以，用 string str;cin>>str; 来接受输入的Push或者Pop在strcmp上是运行不过去的

就可以用char ch[5]; cin>>ch; 来进行字符串的输入

以下是代码

#include <bits/stdc++.h>
using namespace std;
struct node
{
    int data;
   // int layer;              ///存放层数
    node * left;
    node * right;
};
node * root = new node;
int n;
int pre[100];
int in[100];
///  返回的是root根节点的地址
node *CreatTree(int PL,int PR,int IL,int IR)            ///由先序遍历和中序遍历创建一颗二叉树  方法用的是根据先序的左右子树
                                                        ///范围和中序的左右子树的范围
{
    if(PL>PR)        ///说明孩子树已经不存在了
        return NULL;
    int k=0;
    node* root=new node;
    root->data=pre[PL];
    for(k=IL; k<=IR; k++)
    {
        if(pre[PL] == in[k])                ///说明找到了根节点
            break;
    }
    int numLeft=k-IL;
    root->left=CreatTree(PL+1,PL+numLeft,IL,k-1);                ///返回的是左子树的地址
    root->right=CreatTree(PL+numLeft+1,PR,k+1,IR);                    ///返回的是右子树的地址
    return root;
}

int cnt=0;                    ///cnt用来进行控制空格的输出
void postorder(node * root)                ///后序遍历
{
    if(root==NULL)
        return ;

    postorder(root->left);
    postorder(root->right);
    cout<<root->data;
    if(cnt<n)
        cout<<" ";
    cnt++;
}

void f()                    ///题目上要求的输入
{
    stack<int> st;                ///用堆栈来对题目中的Push和Pop进行模拟
    cin>>n;
    int pk=0;
    int ik=0;
    for(int i=0; i<2*n; i++)
    {
        char str[5];
        cin>>str;
        if(strcmp(str,"Push")==0)
        {
            int num;
            cin>>num;
            st.push(num);
            pre[pk++]=num;
        }
        else
        {
            in[ik++]=st.top();
            st.pop();
        }
    }
}

int main()
{
    f();
    root=CreatTree(0,n-1,0,n-1);
    postorder(root);
    return 0;
}