1043 Is It a Binary Search Tree (25分)
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 (≤1000). 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

思路：
根據題目數據先建立二叉搜索樹；
進行判斷：
1.是否是先序遍歷結果。
2.是否是鏡像的先序遍歷結果。

#include<iostream> #include<vector> using namespace std; struct node {int data;node* l;node* r; }; vector<int>pre, mirror_pre, post; void Insert(struct node*& root, int a)//&root，c++引用； {if (root == nullptr){root = new node();root->data = a;return;}if (a < root->data)Insert(root->l, a);else Insert(root->r, a);return; } struct node* create_tree(vector<int>v) {struct node* root = nullptr;for (int i = 0; i < v.size(); i++)//二叉搜索樹建設{Insert(root, v[i]);}return root; } void preorder(struct node* root)//先序遍歷，結果保存在pre數組中 {if (root){pre.push_back(root->data);preorder(root->l);preorder(root->r);} } void postorder(struct node* root)//后序遍歷，結果保存在post數組 {if (root){postorder(root->l);postorder(root->r);post.push_back(root->data);} } void mirror_preorder(struct node* root)//鏡像先序遍歷，結果保存在mirror_pre數組中 {if (root){mirror_pre.push_back(root->data);mirror_preorder(root->r);mirror_preorder(root->l);} } void mirror_postorder(struct node* root)//鏡像后序遍歷，結果保存在post數組 {if (root){mirror_postorder(root->r);mirror_postorder(root->l);post.push_back(root->data);} } int main() {int n;cin >> n;vector<int>v(n);//v存儲給定的數組for (int i = 0; i < n; i++){cin >> v[i];}struct node* root = create_tree(v);//建樹preorder(root);//先進行先序遍歷，保存結果。mirror_preorder(root);//保存鏡像先序遍歷值if (v == pre)//如果先序遍歷數組和題目給定的判別數組相同{postorder(root);//保存后續遍歷數組cout << "YES" << endl;for (int i = 0; i < post.size(); i++){if (i != 0)cout << " ";cout << post[i];}}else if (v == mirror_pre)//如果鏡像先序遍歷數組和題目給定的判別數組相同{mirror_postorder(root);//保存鏡像的后序遍歷數組cout << "YES" << endl;for (int i = 0; i < post.size(); i++){if (i != 0)cout << " ";cout << post[i];}}elsecout << "NO"; }

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