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1043 Is It a Binary Search Tree (25分)(树的插入)

时间:2019-12-12 19:49:32      阅读:100      评论:0      收藏:0      [点我收藏+]

标签:element   -o   extra   one   cas   oid   text   sample   插入   

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

  • The left subtree of a node contains only nodes with keys less than the node‘s key.
  • The right subtree of a node contains only nodes with keys greater than or equal to the node‘s key.
  • Both the left and right subtrees must also be binary search trees.

If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST.

Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤). Then N integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in a line YES if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or NO if not. Then if the answer is YES, print in the next line the postorder traversal sequence of that tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:

7
8 6 5 7 10 8 11

Sample Output 1:

YES
5 7 6 8 11 10 8

Sample Input 2:

7
8 10 11 8 6 7 5

Sample Output 2:

YES
11 8 10 7 5 6 8

Sample Input 3:

7
8 6 8 5 10 9 11

Sample Output 3:

NO
题目分析:树的插入 因为给的是先序遍历 所以每次插入时 如果插入的是左子树 那么它的父亲的右节点必然为空 不然就插入失败 对于对称的情况也是类似
因此 我们只需要知道在插入时 每个节点左右子树是否存在就可
技术图片
  1 #define _CRT_SECURE_NO_WARNINGS
  2 #include <climits>
  3 #include<iostream>
  4 #include<vector>
  5 #include<queue>
  6 #include<map>
  7 #include<set>
  8 #include<stack>
  9 #include<algorithm>
 10 #include<string>
 11 #include<cmath>
 12 using namespace std;
 13 typedef struct Node* PtrToNode;
 14 vector<int> V;
 15 struct Node{
 16     int data;
 17     PtrToNode Left, Right;
 18     bool LE, RI;
 19 };
 20 bool Insert(PtrToNode&T,int Element)
 21 {
 22     if (!T){
 23         T = new Node;
 24         T->data = Element;
 25         T->Left = NULL;
 26         T->Right = NULL;
 27         T->LE = false;
 28         T->RI = false;
 29     }
 30     else
 31         if (Element < T->data){
 32             if (T->RI)
 33                 return false;
 34             else{
 35                 T->LE = true;
 36                 return Insert(T->Left, Element);    
 37             }
 38         }
 39         else{
 40             T->RI = true;
 41             return Insert(T->Right, Element);
 42         }
 43     return true;
 44 }
 45 bool InsertR(PtrToNode& T, int Element)
 46 {
 47     if (!T) {
 48         T = new Node;
 49         T->data = Element;
 50         T->Left = NULL;
 51         T->Right = NULL;
 52         T->LE = false;
 53         T->RI = false;
 54     }
 55     else
 56         if (Element >=T->data) {
 57             if (T->RI)
 58                 return false;
 59             else {
 60                 T->LE = true;
 61                 return InsertR(T->Left, Element);
 62             }
 63         }
 64         else {
 65             T->RI = true;
 66             return InsertR(T->Right, Element);
 67         }
 68     return true;
 69 }
 70 void PostOrder(PtrToNode T)
 71 {
 72     if(T)
 73     {
 74         PostOrder(T->Left);
 75         PostOrder(T->Right);
 76         V.push_back(T->data);
 77     }
 78 }
 79 int main()
 80 {
 81     int N;
 82     cin >> N;
 83     int num;
 84     PtrToNode TL = NULL, TR = NULL;
 85     bool flagL,flagR;
 86     flagL = flagR = true;
 87     for (int i = 0; i < N; i++)
 88     {
 89         cin >> num;
 90         if(flagL)flagL = Insert(TL, num);
 91         if(flagR)flagR = InsertR(TR, num);
 92         if (!flagL&&!flagR)
 93             break;
 94     }
 95     if (flagL||flagR)
 96     {
 97         cout << "YES"<<endl;
 98         if (flagL)
 99             PostOrder(TL);
100         else
101             PostOrder(TR);
102         for (auto it = V.begin(); it != (V.end() - 1); it++)
103             cout << *it << " ";
104         cout << *(V.end() - 1);
105     }
106     else
107         cout << "NO";
108 
109 }
View Code

 

1043 Is It a Binary Search Tree (25分)(树的插入)

标签:element   -o   extra   one   cas   oid   text   sample   插入   

原文地址:https://www.cnblogs.com/57one/p/12031038.html

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