大致题意就是给出一个包含N个整数的序列,构造出一个二叉查找树,并判断这个序列是否与该二叉查找树的先序序列或者镜像二叉查找树的先序序列一样,如果一样,就输出该二叉查找树的后序序列,否则输出NO。
1 #include<iostream> 2 #include<vector> 3 using namespace std; 4 5 struct Node { 6 int data; 7 Node* lchild; 8 Node* rchild; 9 } ; 10 11 vector<int> origin,pre,post; 12 void insert_BST(Node*& root,int data) { //建立BST 13 if(root == NULL) { 14 root = new Node; 15 root->data = data; 16 root->lchild = root->rchild = NULL; 17 return ; 18 } 19 if(data < root->data) insert_BST(root->lchild,data); 20 else insert_BST(root->rchild,data); 21 } 22 void insert_MIBST(Node*& root,int data) {//建立镜像BST 23 if(root == NULL) { 24 root = new Node; 25 root->data = data; 26 root->lchild = root->rchild = NULL; 27 return ; 28 } 29 //不同点 30 if(data >= root->data) insert_MIBST(root->lchild,data); 31 else insert_MIBST(root->rchild,data); 32 } 33 34 void preorder(Node* root) { //先序遍历 35 if(root == NULL) return; 36 pre.push_back(root->data); 37 preorder(root->lchild); 38 preorder(root->rchild); 39 } 40 void postorder(Node* root) {//后序遍历 41 if(root == NULL) return; 42 postorder(root->lchild); 43 postorder(root->rchild); 44 post.push_back(root->data); 45 } 46 int main() { 47 int n,data; 48 cin>>n; 49 Node* rootBST = NULL; //定义BST头结点 50 Node* rootMIBST = NULL;//定义MIBST头结点 51 for(int i = 0; i < n; ++i) { 52 cin>>data; 53 origin.push_back(data); 54 insert_BST(rootBST,data);//构造BST 55 insert_MIBST(rootMIBST,data);//构造镜像BST 56 } 57 preorder(rootBST);//对BST进行先序遍历 58 if(origin == pre) { 59 printf("YES\n"); 60 postorder(rootBST); 61 for(int i = 0; i < post.size(); ++i) { 62 if(i > 0) printf(" "); 63 printf("%d",post[i]); 64 } 65 } else { 66 pre.clear(); 67 preorder(rootMIBST); 68 if(origin == pre) { 69 printf("YES\n"); 70 postorder(rootMIBST); 71 for(int i = 0; i < post.size(); ++i) { 72 if(i > 0) printf(" "); 73 printf("%d",post[i]); 74 } 75 } else 76 printf("NO"); 77 } 78 return 0; 79 }
1043 Is It a Binary Search Tree
原文:https://www.cnblogs.com/keep23456/p/12403195.html