PAT Advanced 1086 Tree Traversals Again (25) [树的遍历]

Posted 2021-03-09 houzm

题目

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.
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

题目分析

已知非递归中序（借助栈）遍历树的操作流程，求后序序列
注：操作流程中push操作依次组成前序序列，又可以借助push和pop操作得到中序序列，题目即转换为中序+前序->后序

解题思路

思路 01

1. 中序+前序建树
2. 递归后序遍历树

思路 02（最优）

1. 中序+前序直接转后序序列

知识点

1. while(~scanf("%s",s)) {} //等价于scanf("%s",s)!=EOF

   两者作用是相同的
   ~是按位取反
   scanf的返回值是输入值的个数
   如果没有输入值就是返回-1
   -1按位取反结果是0
   while(~scanf("%d", &n))就是当没有输入的时候退出循环
   EOF，为End Of File的缩写，通常在文本的最后存在此字符表示资料结束。
   EOF 的值通常为 -1

Code

Code 01

#include <iostream>
#include <vector>
#include <stack>
using namespace std;
vector<int> pre,in;
int index,n;
struct node {
    int data;
    node * left;
    node * right;
};
node * create(int preL,int preR,int inL,int inR) {
    if(preL>preR)return NULL;
    node * now = new node;
    now->data=pre[preL];
    int k=inL;
    while(k<inR&&in[k]!=pre[preL])k++;
    now->left=create(preL+1, preL+(k-inL), inL, k-1);
    now->right=create(preL+(k-inL)+1, preR, k+1, inR);
    return now;
}
void postOrder(node * root) {
    if(root==NULL)return;
    postOrder(root->left);
    postOrder(root->right);
    printf("%d",root->data);
    if(++index<n)printf(" ");
}
int main(int argc,char * argv[]) {
    int id;
    string s;
    stack<int> sc;
    scanf("%d",&n);
    int len=n<<1;
    for(int i=0; i<len; i++) {
        cin>>s;
        if(s=="Push") {
            scanf("%d",&id);
            pre.push_back(id);//preorder
            sc.push(id);
        }
        if(s=="Pop") {
            in.push_back(sc.top());
            sc.pop();
        }
    }
    node * root = create(0,n-1,0,n-1);
    postOrder(root);
    return 0;
}

Code 02（最优）

#include <iostream>
#include <vector>
#include <stack>
#include <cstring>
using namespace std;
vector<int> pre,in,post;
int n;
void postOrder(int preL,int preR,int inL,int inR){
    if(inL>inR)return;// preL>preR也可以 
    int k=inL;
    while(k<inR&&in[k]!=pre[preL])k++;
    postOrder(preL+1, preL+(k-inL), inL, k-1);//先存放左子节点 
    postOrder(preL+(k-inL)+1, preR, k+1, inR);//后存放右子节点 
    post.push_back(pre[preL]); //再存放父节点 
} 
int main(int argc,char * argv[]) {
    int id;
    char s[5];
    stack<int> sc;
    scanf("%d",&n);
    while(~scanf("%s",s)) { //等价于scanf("%s",s)!=EOF 
        if(strlen(s)==4) {
            //Push
            scanf("%d",&id);
            pre.push_back(id);
            sc.push(id);
        } else {
            //Pop
            in.push_back(sc.top());
            sc.pop();
        }
    }
    postOrder(0,n-1,0,n-1);
    for(int i=0;i<post.size();i++){
        printf("%d",post[i]);
        if(i!=post.size()-1)printf(" ");
    }
    return 0;
}