PAT 1043. Is It a Binary Search Tree

1043. Is It a Binary Search Tree (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

78 6 5 7 10 8 11

Sample Output 1:

YES5 7 6 8 11 10 8

Sample Input 2:

78 10 11 8 6 7 5

Sample Output 2:

YES11 8 10 7 5 6 8

Sample Input 3:

78 6 8 5 10 9 11

Sample Output 3:

NO

可以直接在建bst的途中确定yes no

对于bst，前序遍历，一旦遍历了某个右节点，那么以后绝对不会再遍历对应的左子树了

#include <cstdio>#include <cstdlib>struct Node{int val;Node *left, *right;// 如果强行访问，说明不是前序bool banLeft, banRight;};bool bMirror;bool Fill(Node* root, Node* newNode){if (newNode->val < root->val){if (root->banLeft)return false;// 对于mirror// 一旦访问了左侧，说明不会再访问右侧if(bMirror)root->banRight = true;if (root->left == nullptr){root->left = newNode;}else{return Fill(root->left, newNode);}}else{if (root->banRight)return false;// 对于正常bst// 一旦访问了右侧，说明不会再访问左侧if (!bMirror)root->banLeft = true;if (root->right == nullptr){root->right = newNode;}else{return Fill(root->right, newNode);}}return true;}void PrintPostOrder(Node* p){// 左右中// 对于mirror反过来if (bMirror){if (p->right)PrintPostOrder(p->right);if (p->left)PrintPostOrder(p->left);}else{if (p->left)PrintPostOrder(p->left);if (p->right)PrintPostOrder(p->right);}static bool bFirst = true;if (bFirst)bFirst = false;elseprintf(" ");printf("%d", p->val);}int main(){int N;scanf("%d", &N);if (N == 1){int tmp;scanf("%d", &tmp);printf("YES\n%d", tmp);return 0;}// 前两个手动，确定bst还是镜像int tmp;scanf("%d", &tmp);Node* pRoot = new Node{ tmp, nullptr, nullptr,false,false };scanf("%d", &tmp);Node* pFirstChild = new Node{ tmp, nullptr, nullptr,false,false };if (pFirstChild->val < pRoot->val)bMirror = false;elsebMirror = true;Fill(pRoot, pFirstChild);for (int i = 2; i < N; ++i){scanf("%d", &tmp);if (!Fill(pRoot, new Node{ tmp,nullptr,nullptr,false,false })){// 失败了，不是前序printf("NO");return 0;}}// 是前序，且造好了bstprintf("YES\n");// 输出后序PrintPostOrder(pRoot);system("pause");return 0;}