标签：osi efi mib board nbsp input seq als bsp
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:
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.
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.
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.
7
8 6 5 7 10 8 11
YES
5 7 6 8 11 10 8
7
8 10 11 8 6 7 5
YES
11 8 10 7 5 6 8
7
8 6 8 5 10 9 11
NO题意：
给出一个数字序列，问这个序列是不是一棵BST或者Mirror BST前序遍历的结果，如果是则输出对应树的后序遍历结果，如果不是则输出NO。
思路：
施工中………………
Code：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int n; 6 vector<int> v; 7 vector<int> post; 8 9 typedef struct Node* node; 10 11 struct Node { 12 int val; 13 node left; 14 node right; 15 Node(int v) { 16 val = v; 17 left = NULL; 18 right = NULL; 19 } 20 }; 21 22 bool isBST() { 23 bool found1 = false; 24 for (int i = 1; i < n; ++i) { 25 if (v[i] >= v[0]) found1 = true; 26 if (found1 && v[i] < v[0]) return false; 27 } 28 return true; 29 } 30 31 bool isMIBST() { 32 bool found2 = false; 33 for (int i = 1; i < n; ++i) { 34 if (v[i] < v[0]) found2 = true; 35 if (found2 && v[i] >= v[0]) return false; 36 } 37 return true; 38 } 39 40 node buildTree(int s, int e, bool isBST) { 41 if (s > e) return NULL; 42 node root = new Node(v[s]); 43 int pos = -1; 44 if (isBST) { 45 for (int i = s + 1; i <= e; ++i) 46 if (v[i] >= v[s]) { 47 pos = i; 48 break; 49 } 50 } else { 51 for (int i = s + 1; i <= e; ++i) 52 if (v[i] < v[s]) { 53 pos = i; 54 break; 55 } 56 } 57 if (pos == -1) return root; 58 root->left = buildTree(s + 1, pos - 1, isBST); 59 root->right = buildTree(pos, e, isBST); 60 return root; 61 } 62 63 void postOrder(node root) { 64 if (root == NULL) return; 65 postOrder(root->left); 66 postOrder(root->right); 67 post.push_back(root->val); 68 } 69 70 int main() { 71 cin >> n; 72 v.resize(n + 1); 73 for (int i = 0; i < n; ++i) cin >> v[i]; 74 node root; 75 if (isBST()) { 76 cout << "YES" << endl; 77 root = buildTree(0, n - 1, true); 78 } else if (isMIBST()) { 79 cout << "YES" << endl; 80 root = buildTree(0, n - 1, false); 81 } else { 82 cout << "NO" << endl; 83 return 0; 84 } 85 postOrder(root); 86 bool isFirst = true; 87 for (int i : post) { 88 if (isFirst) { 89 cout << i; 90 isFirst = false; 91 } else { 92 cout << " " << i; 93 } 94 } 95 cout << endl; 96 return 0; 97 }
注意：
上面的代码有三组数据不能通过，有时间再更正…………
1043 Is It a Binary Search Tree
标签：osi efi mib board nbsp input seq als bsp
原文地址：https://www.cnblogs.com/ruruozhenhao/p/12865952.html
