PAT Advanced 1043 Is It a Binary Search Tree (25) [⼆叉查找树BST]

题目

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The lef 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 lef and right subtrees must also be binary search trees. If we swap the lef 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

题目分析

已知二叉查找树节点序列，判断是其前序序列还是其镜像树的前序序列，并打印相应树的后序序列

解题思路

思路 01

1. 输入测试数据时，分别建树和建镜像树
2. 先用树的先序序列与原测试序列对比，若同即输出YES，若不同再用镜像树先序序列对比，若同输出YES，不同则NO
3. 若为YES，打印相应后序序列

思路 02（最优、难理解）

1. 输入测试数据时，建树
2. 根据二叉查找树的性质（大于所有左子树节点，小于所有右子树节点）
   
   2.1 获取后序序列，若后序序列中的结点数与原测试用例结点数相同，即为二叉查找树的先序序列打印YES，若不同，清空，并进性镜像树的后序序列获取
   
   2.2 获取镜像树的后序序列结点数，若与原测试用例结点数相同，即为二叉查找树镜像树的先序序列YES，若不同，打印NO
3. 若为YES，打印相应后序序列

知识点

二叉查找树的前序转后序，无需建树，可根据其性质（大于所有左子树节点，小于所有右子树节点）建树

如前序序列：8 6 5 7 10 8 11

8是根节点

左子树：从6开始往后找小于8的都为8的左子树节点

右子树：从最后一位11开始往前找大于8的都为8的右子树节点

继续递归过程，直到完成建树

Code

Code 01

#include <iostream>
#include <vector>
using namespace std;
struct node {
	int data;
	node * left=NULL;
	node * right=NULL;
	node() {}
	node(int _data):data(_data) {}
};
node * root,* rootM;
void insert(int n, int b) {
	if(root==NULL&&b==0) {
		root = new node(n);
		return;
	}
	if(rootM==NULL&&b==1) {
		rootM = new node(n);
		return;
	}
	node * p;
	if(b==0)p=root;
	else p=rootM;
	while(p!=NULL) {
		if((n<p->data&&b==0)||(n>=p->data&&b==1)) {
			if(p->left==NULL) {
				p->left=new node(n);
				return;
			}
			p=p->left;
		} else if((n>=p->data&&b==0)||(n<p->data&&b==1)) {
			if(p->right==NULL) {
				p->right=new node(n);
				return;
			}
			p=p->right;
		}
	}
}
vector<int> origin,pre,post,preM,postM;
void preOrder(node * nd, int b) {
	if(nd==NULL)return;
	if(b==0)pre.push_back(nd->data);
	else preM.push_back(nd->data);
	preOrder(nd->left,b);
	preOrder(nd->right,b);
}
void postOrder(node * nd, int b) {
	if(nd==NULL)return;
	postOrder(nd->left,b);
	postOrder(nd->right,b);
	if(b==0)post.push_back(nd->data);
	else postM.push_back(nd->data);
}
int main(int argc,char * argv[]) {
	int n,m;
	scanf("%d",&n);
	for(int i=0; i<n; i++) {
		scanf("%d",&m);
		origin.push_back(m);
		insert(m,0);
		insert(m,1);
	}
//	int flag = -1;//0 前序；1 镜像前序；2 NO
	preOrder(root,0);
	preOrder(rootM,1);
	if(pre==origin) {
		postOrder(root,0);
		printf("YES\n");
		for(int i=0; i<post.size(); i++) {
			if(i!=0)printf(" ");
			printf("%d",post[i]);
		}
	} else if(preM==origin) {
		if(preM==origin) {
			postOrder(rootM,1);
			printf("YES\n");
			for(int i=0; i<postM.size(); i++) {
				if(i!=0)printf(" ");
				printf("%d",postM[i]);
			}
		}
	}else{
		printf("NO\n");
	}
	return 0;
}

Code 02（最优、难理解）

#include <iostream>
#include <vector>
using namespace std;
vector<int> pre,post;
bool isMirror;
void getPost(int root, int tail) {
	if(root>tail)return;
	int i=root+1;
	int j=tail;
	if(!isMirror) {
		while(i<=tail&&pre[i]<pre[root])i++;
		while(j>root&&pre[j]>=pre[root])j--;
	} else {
		while(i<=tail&&pre[i]>=pre[root])i++;
		while(j>root&&pre[j]<pre[root])j--;
	}
	if(i-j!=1)return;
	getPost(root+1,j);//左子树
	getPost(i,tail); //右子树
	post.push_back(pre[root]);
}
int main(int argc,char * argv[]) {
	int n,m;
	scanf("%d",&n);
	for(int i=0; i<n; i++) {
		scanf("%d",&m);
		pre.push_back(m);
	}
	getPost(0,n-1);
	if(post.size()!=n) {
		isMirror=true;
		post.clear();
		getPost(0,n-1);
	}
	if(post.size()==n) {
		printf("YES\n%d",post[0]);
		for(int i=1; i<post.size(); i++) {
			printf(" %d",post[i]);
		}
	} else {
		printf("NO\n");
	}
	return 0;
}