【C】二叉查找树(BST)

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:

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

#include<stdio.h>
#include<vector>
using namespace std;
struct node{
	int data;
	node *leftchild,*rightchild;
};
void insert(node* &root,int data){
	if(root==NULL){
		root=new node;
		root->data=data;
		root->leftchild=root->rightchild=NULL;
		return;
	}
	if(data<root->data) insert(root->leftchild,data);//要插入的数小于当前数
	else insert(root->rightchild,data);
}
void preorder(node* root,vector<int> &vi){
	if(root==NULL) return;
	vi.push_back(root->data);
	preorder(root->leftchild,vi);
	preorder(root->rightchild,vi);
}
void preorderMirror(node* root,vector<int> &vi){
	if(root==NULL) return;
	vi.push_back(root->data);
	preorderMirror(root->rightchild,vi);
	preorderMirror(root->leftchild,vi);
}
void postorder(node* root,vector<int> &vi){
	if(root==NULL) return;
	postorder(root->leftchild,vi);
	postorder(root->rightchild,vi);
	vi.push_back(root->data);
}
void postorderMirror(node* root,vector<int> &vi){
	if(root==NULL) return;
	postorderMirror(root->rightchild,vi);
	postorderMirror(root->leftchild,vi);
	vi.push_back(root->data);
}
vector<int> origin,pre,preM,post,postM;
int main(){
	int n,i;
	node* root=NULL;
	scanf("%d",&n);
	for(i=0;i<n;i++){
		int a;
		scanf("%d",&a);
		origin.push_back(a);
		insert(root,a);
	}
	preorder(root,pre);
	preorderMirror(root,preM);
	postorder(root,post);
	postorderMirror(root,postM);
	if(origin==pre){//先序
		printf("YES\n");
		for(i=0;i<post.size();i++){
			printf("%d",post[i]);
			if(i<post.size()-1) printf(" ");
		}
	}
	else if(origin==preM){
		printf("YES\n");
		for(i=0;i<postM.size();i++){
			printf("%d",postM[i]);
			if(i<postM.size()-1) printf(" ");
		}
	}
	else printf("NO");
	return 0;
}

1064. Complete Binary Search Tree (30)

时间限制 100 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.

A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.

Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding complete binary search 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:

10
1 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

//根据给定序列,建一棵完全二叉搜索树 
#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{
	int data;
	int lchild;
	int rchild;
}Node[1010];
int n,len=1;
/*void initial(){
	int i=1;
	while(i*2<=n){
		Node[i].lchild=i*2;
		i*=2;
	}
	i=1;
	while(i*2+1<=n){
		Node[i].rchild=i*2+1;
		i=i*2+1;
	}
}*/
bool cmp(int a,int b){
	return a<b;
}
void create(int data[],int index){
	if(index*2>n){
		Node[index].data=data[len];
		//printf("Node[%d]=%d\n",index,data[len]);
		len++;
		return;
	}
	if(index*2<=n) create(data,index*2);
	Node[index].data=data[len];len++;
	if(index*2+1<=n) create(data,index*2+1);
}
int main(){
	int buf[1010];
	scanf("%d",&n);
	int i;
	for(i=1;i<=n;i++){
		scanf("%d",&buf[i]);
	}
	sort(buf+1,buf+n+1,cmp);
	create(buf,1);
	for(i=1;i<=n-1;i++){
		printf("%d ",Node[i].data);
	}
	printf("%d",Node[n].data);
	return 0;
} 
	

1099. Build A Binary Search Tree (30)

时间限制 100 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.

Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.

《【C】二叉查找树(BST)》

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format “left_index right_index”, provided that the nodes are numbered from 0 to N-1, and 0 is always the root. If one child is missing, then -1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.

Output Specification:

For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

9
1 6
2 3
-1 -1
-1 4
5 -1
-1 -1
7 -1
-1 8
-1 -1
73 45 11 58 82 25 67 38 42

Sample Output:

58 25 82 11 38 67 45 73 42

//先建好这棵树,根据节点序号进行中序遍历
//将给定序列按递增序依次填入中序遍历序列
//输出层次遍历节点的数值
//注意0号总是根
#include<stdio.h>
#include<algorithm>
#include<queue>
using namespace std;
struct node{
	int data;
	int lchild;
	int rchild;
}Node[110];
int n,len=0;
bool cmp(int a,int b){
	return a<b;
}
void inorder(int buf[],int index){
	if(Node[index].lchild!=-1) inorder(buf,Node[index].lchild);
	Node[index].data=buf[len];len++;
	if(Node[index].rchild!=-1) inorder(buf,Node[index].rchild);
}
void levelorder(int root){
	queue<int> q;
	q.push(root);
	int f=1;
	printf("%d",Node[root].data);
	while(!q.empty()){
		int p=q.front();
		q.pop();
		if(f==1) f=0;
		else printf(" %d",Node[p].data);
		if(Node[p].lchild!=-1) q.push(Node[p].lchild);
		if(Node[p].rchild!=-1) q.push(Node[p].rchild);
	}
}

int main(){
	scanf("%d",&n);
	int i;
	int buf[110];
	for(i=0;i<n;i++){
		int a,b;
		scanf("%d %d",&a,&b);
		Node[i].lchild=a;
		Node[i].rchild=b;
	}
	for(i=0;i<n;i++){
		scanf("%d",&buf[i]);
	}
	sort(buf,buf+n,cmp);
	inorder(buf,0);
	levelorder(0);
	return 0;
}
    原文作者:二叉查找树
    原文地址: https://blog.csdn.net/Li_Jiaqian/article/details/79426505
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