浙大 PAT 1043. Is It a Binary Search Tree (25)

1043. Is It a Binary Search Tree (25)

时间限制

400 ms

内存限制

32000 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:

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

这个题目自我感觉描述有问题，题目说如果是yes则输出该树的后序遍历序列。可其实当该序列为其mirror的前序时要输出的序列是该树mirror的后序序列。这个题目我没看出来，但是第二个input给出了这样的情况。

先建树，这没什么好说的，然后各种遍历取得遍历序列比较即可的yes or no。然后yes的时候要输出其后序，当要输出其mirror的后序时有两种方法可供选择，1.直接按右左根遍历该树输出 2.将树通过交换左右子树转化为其mirror然后后序遍历。这两种方法我都试了，都能正常通过，这里用了第二种方法。

#include <vector>
#include <cstdio>
#include <cstdlib>

using namespace std;

const int maxx = 1002;

typedef struct Tree{
	Tree * left;
	Tree * right;
	int ele;
}Tree;

vector<int> righta,rightb;
vector<int> post,seq;

void insert(Tree *&root,int num){
	if(root==NULL){
		root = (Tree *)malloc(sizeof(Tree));
		root->ele = num;
		root->left = NULL;
		root->right = NULL;
		return;
	}

	if(num<root->ele){
		insert(root->left,num);
	}else{
		insert(root->right,num);
	}
}

void preOrder(Tree *root){//前序
	if(root==NULL)return;
	righta.push_back(root->ele);
	preOrder(root->left);
	preOrder(root->right);
}

void ARL(Tree *root){//根右左mirror的前序
	if(root==NULL)return;
	rightb.push_back(root->ele);
	ARL(root->right);
	ARL(root->left);
}

void postOrder(Tree *root){//后序
	if(root==NULL)return;
	postOrder(root->left);
	postOrder(root->right);
	post.push_back(root->ele);
}

void print(vector<int> vc){
	vector<int>::iterator ite = vc.begin();

	bool flg = true;
	for(;ite!=vc.end();++ite){
		if(!flg){
			printf(" %d",*ite);
		}else{
			printf("%d",(*ite));
			flg = false;
		}
	}
	printf("\n");
}

void change(Tree *root){
	if(root==NULL)return;
	Tree *temp = root->left;
	root->left = root->right;
	root->right = temp;
	change(root->left);
	change(root->right);
}

int main(){
	int n,i,num;
	Tree *root = NULL;
	scanf("%d",&n);
	for(i=0;i<n;++i){
		scanf("%d",&num);
		seq.push_back(num);
		insert(root,num);
	}
	
	preOrder(root);
	ARL(root);
	if(righta==seq){
		printf("YES\n");
		post.clear();
		postOrder(root);
		print(post);
	}else if(rightb==seq){
		printf("YES\n");
		change(root);
		post.clear();
		postOrder(root);
		print(post);
	}else{
		printf("NO\n");
	}
	return 0;
}