PAT 1086 Tree Traversals Again[中序转后序][难]

1086 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

题目大意：二叉树的中根遍历，可以通过栈来实现，那么现在给出一棵二叉树的中根遍历操作，要求输出后根遍历结果。

//完全可以通过输入来确定这棵二叉树的中根遍历，即已知中根遍历求后根遍历。但是我不会啊。

代码转自：https://www.liuchuo.net/archives/2168

#include <cstdio>
#include <vector>
#include <stack>
#include <cstring>
using namespace std;
vector<int> pre, in, post,value;
void postorder(int root, int start, int end) {
    if (start > end) return;
    int i = start;
    while (i < end && in[i] != pre[root]) {//中序遍历序列中存的节点的id,唯一的！
        i++;
        printf("%d %d\n" ,in[i],pre[root]);
    }
    postorder(root + , start, i - );
    //左子树共有i-start+1个节点。
    postorder(root +  + i - start, i + , end);
    post.push_back(pre[root]);
}
int main() {
    int n;
    scanf("%d", &n);
    char str[];
    stack<int> s;
    int key=;
    while (~scanf("%s", str)) {
        if (strlen(str) == ) {
            int num;
            scanf("%d", &num);
            value.push_back(num);
            pre.push_back(key);//对应num有一个序号，从0开始。

            s.push(key++);
        } else {
            in.push_back(s.top());//现在存了中序遍历
            //存的是id对应的序号（为了防止重复呢。）
            s.pop();
        }
    }

    postorder(, , n - );
    printf("\n");
    printf("%d", value[post[]]);
    for (int i = ; i < n; i++)
        printf(" %d",value[post[i]]);
    return ;
}

//这个代码简直太难了，看了好几遍都理解不了那个中序转后序的，气死了。

//这个明天还要搜一下别的题解，简直气死我了。

//更要重点掌握一套，二叉树的各种访问序列转换方法。

2018-11-17更——————

我的AC：

#include <iostream>
#include <cstdio>
#include <vector>
#include<stack>

using namespace std;
vector<int> in,pre,post;
void postOrder(int inL,int inR,int preL,int preR){
    if(inL>inR)return ;
   // int i=0;//标识中根遍历中的根节点下标
    int i=;
    while(in[i]!=pre[preL])i++;
    //遍历左右子树
    postOrder(inL,i-,preL+,preR+i-inL);
    postOrder(i+,inR,preL+i-inL+,preR);
    post.push_back(in[i]);
}
int main()
{
    //push的顺序就是前序，弹出的顺序就是中序。
    int n,id;
    cin>>n;
    string s;
    stack<int> tree;
    for(int i=;i<*n;i++){
        cin>>s;
        if(s[]=='u'){
            cin>>id;
            tree.push(id);
            pre.push_back(id);//前序遍历放进来。
        }else{
            int temp=tree.top();
            tree.pop();
            in.push_back(temp);
        }
    }
   // cout<<pre.size();
    postOrder(,n-,,n-);
    for(int i=;i<n;i++){
        cout<<post[i];
        if(i!=n-)cout<<" ";
    }
    return ;
}

//在牛客网上通不过，说内存超限，通过率为0，因为递归层数太深？

遇到的问题：

1.postOrder函数，作为递归出口应该是in的左右去判断，如果是pre的，则不会输出结果

2.在postOrder的while循环中，i可以从0开始判断。

3.柳神的代码考虑了key不唯一的情况，但是我没考虑，而且PAT上应该也没考虑，否则就不会AC了。

4.关于这个key的问题，是应该考虑一下不唯一的情况的，因为题目里并没有说。