PAT A1043 Is It a Binary Search Tree (25分)

港控/mmm° 2023-02-20 07:52 9阅读 0赞

[A1043 Is It a Binary Search Tree (25分)][A1043 Is It a Binary Search Tree _25]  
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:

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3_size_16_color_FFFFFF_t_70]  
代码一  
[来自柳诺][Link 1]  
题目大意：给定一个整数键值序列，现请你编写程序，判断这是否是对一棵二叉搜索树或其镜像进行前序遍历的结果。  
分析：假设它是二叉搜索树，一开始isMirror为false，根据二叉搜索树的性质将已知的前序转换为后序，转换过程中，如果发现最后输出的后序数组长度不为n，那就设isMirror为true，然后清空后序数组，重新再转换一次（根据镜面二叉搜索树的性质），如果依旧转换后数组大小不等于n，就输出NO否则输出YES

#include <cstdio>
    #include <vector>
    using namespace std;
    bool isMirror;
    vector<int> pre, post;
    void getpost(int root, int tail) { 
        if(root > tail) return ;
        int i = root + 1, j = tail;
        if(!isMirror) { 
            while(i <= tail && pre[root] > pre[i]) i++;
            while(j > root && pre[root] <= pre[j]) j--;
        } else { 
            while(i <= tail && pre[root] <= pre[i]) i++;
            while(j > root && pre[root] > pre[j]) j--;
        }
        if(i - j != 1) return ;
        getpost(root + 1, j);
        getpost(i, tail);
        post.push_back(pre[root]);
    }
    int main() { 
        int n;
        scanf("%d", &n);
        pre.resize(n);
        for(int i = 0; i < n; i++)
            scanf("%d", &pre[i]);
        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 < n; i++)
                printf(" %d", post[i]);
        } else { 
            printf("NO");
        }
        return 0;
    }

代码二

//PATA1043-Is-It-a-Binary-Search-Tree
    #include <iostream>
    #include <cstdio>
    #include <vector>
    using namespace std;
    struct node{ 
    	int data;
    	node *left,*right;
    };
    void insert(node *&root,int data){ 
    	if(root == NULL){ //到达空结点即为需要插入的位置 
    		root = new node;
    		root->data = data;
    		root->left = root->right = NULL;
    		return;
    	}
    	if(data < root->data)	insert(root->left,data);
    	else insert(root->right,data);
    }
    void preOrder(node *root,vector<int> &vi){ //先序遍历，结果存在vi 
    	 if(root == NULL)	return;
    	 vi.push_back(root->data);
    	 preOrder(root->left,vi);
    	 preOrder(root->right,vi);
    }
    //镜像树先序遍历，结果存放于vi
    void preOrderMirror(node *root,vector<int> &vi){ 
    	if(root == NULL)	return;
    	vi.push_back(root->data);
    	preOrderMirror(root->right,vi);
    	preOrderMirror(root->left,vi);
    } 
    void postOrder(node *root,vector<int> &vi){ //后序遍历，结果存放于vi中
    	if(root == NULL)	return;
    	postOrder(root->left,vi);
    	postOrder(root->right,vi);
    	vi.push_back(root->data); 
    }
    //镜像树后序遍历，结果存放于vi
    void postOrderMirror(node *root,vector<int> &vi){ 
    	if(root == NULL)	return;
    	postOrderMirror(root->right,vi);
    	postOrderMirror(root->left,vi);
    	vi.push_back(root->data);
    } 
    //origin存放初始序列
    //pre、post为先序和后序，preM、postM为镜像树先序、后序
    vector<int> origin,pre,preM,post,postM;
     
    int main(){ 
    	int n,data;
    	node *root = NULL;//定义头结点
    	scanf("%d",&n);//输入结点个数
    	for(int i = 0;i < n;i++){ 
    		scanf("%d",&data);
    		origin.push_back(data);//将数据加入origin
    		insert(root,data); 
    	} 
    	preOrder(root,pre);//求先序
    	preOrderMirror(root,preM);//求镜像树先序
    	postOrder(root,post);//求后序
    	postOrderMirror(root,postM);//求镜像树后序
    	if(origin == pre){ //初始序列等于先序序列
    		printf("YES\n");
    		for(int i = 0;i < post.size();i++){ 
    			printf("%d",post[i]);
    			if(i < post.size() - 1)	printf(" ");
    		} 
    	} 
    	else if(origin == preM){ //初始序列等于镜像树先序序列
    		printf("YES\n");
    		for(int i = 0;i < postM.size();i++){ 
    			printf("%d",postM[i]);
    			if(i < postM.size() - 1)	printf(" ");
    		} 
    	}
    	else{ 
    		printf("NO\n");
    	}
    	return 0;
    }

#include<bits/stdc++.h>
    using namespace std;
    int n,k=0;
    vector<int> vi;
    bool flag=true,isTree=true;
    struct node{ 
    	int val;
    	node *lc,*rc;
    };
    node *root = NULL;
    node *createa(int le,int end){ 
    	if(le>end) return NULL;
        node *root= new node;
        root->val=vi[le];
        int i;
        for(i=le+1;i<=end;i++){ 
        	if(vi[i]<vi[le]) break;
    	}
    	for(int j=i;j<=end;j++){ 
    		if(vi[j]>=vi[le] ) { 
    			isTree=false;
    			return NULL;
    		}
    	}
        root->lc=createa(le+1,i-1);
        root->rc=createa(i,end);
        return root;
    }
    node *createb(int le,int end){ 
    	if(le>end) return NULL;
        node *root= new node;
        root->val=vi[le];
        int i;
        for(i=le+1;i<=end;i++){ 
        	if(vi[i]>=vi[le])break;	
    	}
    	for(int j=i;j<=end;j++){ 
    		if(vi[j]<vi[le]) { 
    			isTree=false;
    			return NULL;
    		}
    	}
        root->lc=createb(le+1,i-1);
        root->rc=createb(i,end);
        return root;
    }
    void printfTree(node *root){ 
    	if(root==NULL)return;
    	printfTree(root->lc);
    	printfTree(root->rc);
    	if(flag){ 
    		printf("%d",root->val);
    		flag=false;
    	}else printf(" %d",root->val);
    } 
    int main(){ 
    	cin>>n;
    	vi.resize(n+1);
    	for(int i=0;i<n;i++)cin>>vi[i];
    	if(vi[0]>vi[1]){ 
    		root=createb(0,n-1);
    	}else root=createa(0,n-1);
    	if(isTree){ 
    		cout<<"YES"<<endl;
    		printfTree(root);
    	}else cout<<"NO"; 
    	return 0;
    }

[A1043 Is It a Binary Search Tree _25]: https://pintia.cn/problem-sets/994805342720868352/problems/994805440976633856
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3_size_16_color_FFFFFF_t_70]: https://img-blog.csdnimg.cn/20200625110801908.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3,size_16,color_FFFFFF,t_70
[Link 1]: https://www.liuchuo.net/archives/2153