Is It a Binary Search Tree

本文详细介绍了二叉树的前序、中序、后序遍历算法,并通过一个具体的例子展示了如何根据给定的节点插入顺序构建二叉树。此外,还实现了四种不同的遍历方式:前序遍历、前序逆序遍历、后序遍历和后序逆序遍历。

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#include<iostream>
#include<vector>
using namespace std;
int N;
vector<int> origin,post,postM,pre,preM;
struct node{
	int val;
	struct node *left,*right;
}; 
void insert(node* &root,int val){
	if(root==NULL){
		root=new node();
		root->val=val; 
		root->left=root->right=NULL;
	}else if(root->val<=val){
		insert(root->right,val);
	}else
		insert(root->left,val); 
}
void preOrder(node* root){
	if(root==NULL) return;
	pre.push_back(root->val);
	preOrder(root->left);
	preOrder(root->right); 
}
void preMOrder(node* root){
	if(root==NULL) return;
	preM.push_back(root->val);
	preMOrder(root->right);
	preMOrder(root->left); 
}
void postOrder(node* root){
	if(root==NULL) return;
	postOrder(root->left);
	postOrder(root->right);
	post.push_back(root->val);
}
void postMOrder(node *root){
	if(root==NULL) return;
	postMOrder(root->right);
	postMOrder(root->left);
	postM.push_back(root->val);
}
int main(){
	node* root=NULL;
	scanf("%d",&N);
	origin.resize(N);
	for(int i=0;i<N;i++){
		scanf("%d",&origin[i]);
		insert(root,origin[i]);
	} 
	preOrder(root);
	preMOrder(root);
	if(pre==origin){
		printf("YES\n");
		postOrder(root);
		for(int i=0;i<post.size();i++){
			if(i!=0)
				printf(" ");
			printf("%d",post[i]);
		}
	}else if(preM==origin){
		printf("YES\n");
		postMOrder(root);
		for(int i=0;i<postM.size();i++){
			if(i!=0)
				printf(" ");
			printf("%d",postM[i]);
		}
	}else{
		printf("NO\n");
	}
}

 

【Solution】 To convert a binary search tree into a sorted circular doubly linked list, we can use the following steps: 1. Inorder traversal of the binary search tree to get the elements in sorted order. 2. Create a doubly linked list and add the elements from the inorder traversal to it. 3. Make the list circular by connecting the head and tail nodes. 4. Return the head node of the circular doubly linked list. Here's the Python code for the solution: ``` class Node: def __init__(self, val): self.val = val self.prev = None self.next = None def tree_to_doubly_list(root): if not root: return None stack = [] cur = root head = None prev = None while cur or stack: while cur: stack.append(cur) cur = cur.left cur = stack.pop() if not head: head = cur if prev: prev.right = cur cur.left = prev prev = cur cur = cur.right head.left = prev prev.right = head return head ``` To verify the accuracy of the code, we can use the following test cases: ``` # Test case 1 # Input: [4,2,5,1,3] # Output: # Binary search tree: # 4 # / \ # 2 5 # / \ # 1 3 # Doubly linked list: 1 <-> 2 <-> 3 <-> 4 <-> 5 # Doubly linked list in reverse order: 5 <-> 4 <-> 3 <-> 2 <-> 1 root = Node(4) root.left = Node(2) root.right = Node(5) root.left.left = Node(1) root.left.right = Node(3) head = tree_to_doubly_list(root) print("Binary search tree:") print_tree(root) print("Doubly linked list:") print_list(head) print("Doubly linked list in reverse order:") print_list_reverse(head) # Test case 2 # Input: [2,1,3] # Output: # Binary search tree: # 2 # / \ # 1 3 # Doubly linked list: 1 <-> 2 <-> 3 # Doubly linked list in reverse order: 3 <-> 2 <-> 1 root = Node(2) root.left = Node(1) root.right = Node(3) head = tree_to_doubly_list(root) print("Binary search tree:") print_tree(root) print("Doubly linked list:") print_list(head) print("Doubly linked list in reverse order:") print_list_reverse(head) ``` The output of the test cases should match the expected output as commented in the code.
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