1043. Is It a Binary Search Tree (25)

本文介绍了一种使用C++实现的二叉树构建方法,并详细展示了如何通过递归方式完成不同顺序的树节点遍历,包括先序、中序、后序遍历等。此外还提供了释放树内存的方法。

考察建立二叉树,以及树的遍历

#include<iostream>
#include<vector>
#include<queue>

typedef struct Node
{
	int key;
	Node* left;
	Node* right;
	Node(Node* _l = NULL, Node* _r = NULL, int _k = -1)
		:left(_l), right(_r),key(_k){};
}Node;

void InsertTree(Node* &tree, int key)
{
	Node** p = &tree;
	while((*p) != NULL)
	{
		if( key < (*p)->key )
			p = &(*p)->left;
		else p = &(*p)->right;
	}
	(*p) = new Node;
	(*p)->key = key;
}
void ReleaseTree(Node* tree)
{
	std::queue<Node*> q;
	q.push(tree);
	while(!q.empty())
	{
		Node* p = q.front();
		q.pop();
		if(p)
		{
			q.push(p->left);
			q.push(p->right);
			delete p;
		}
	}
}
void TravelRootLR(Node* tree, std::vector<int>& v)
{
	if(tree == NULL) return;
	v.push_back(tree->key);
	TravelRootLR(tree->left, v);
	TravelRootLR(tree->right, v);
}
void TravelRootRL(Node* tree, std::vector<int>& v)
{
	if(tree == NULL) return;
	v.push_back(tree->key);
	TravelRootRL(tree->right, v);
	TravelRootRL(tree->left, v);
}
void TravelLRRoot(Node* tree, std::vector<int>& v)
{
	if(tree == NULL) return;
	TravelLRRoot(tree->left, v);
	TravelLRRoot(tree->right, v);
	v.push_back(tree->key);
}
void TravelRLRoot(Node* tree, std::vector<int>& v)
{
	if(tree == NULL) return;
	TravelRLRoot(tree->right, v);
	TravelRLRoot(tree->left, v);
	v.push_back(tree->key);
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		std::vector<int> inputVec(n);
		for(int i = 0; i < n; ++i)
			scanf("%d",&inputVec[i]);
		//build tree
		Node* tree = NULL;
		for(int i = 0; i < n; ++i)
			InsertTree(tree, inputVec[i]);
		//get RootLR & RootRL & RLRoot
		std::vector<int> t1, t2, t3;
		TravelRootLR(tree, t1);
		TravelRootRL(tree, t2);
		
		//judge
		//int size = (int)t3.size();
		//if( IsSame(inputVec, t1) || IsSame(inputVec, t2) )
		if(inputVec == t1)
		{
			printf("YES\n");
			TravelLRRoot(tree, t3);
			int size = (int)t3.size();
			for(int i = 0; i < size-1; ++i)
				printf("%d ", t3[i]);
			printf("%d\n", t3[size-1]);
		}
		else if(inputVec == t2)
		{
			printf("YES\n");
			TravelRLRoot(tree, t3);
			int size = (int)t3.size();
			for(int i = 0; i < size-1; ++i)
				printf("%d ", t3[i]);
			printf("%d\n", t3[size-1]);
		}
		else 
		{
			printf("NO\n");
		}
		ReleaseTree(tree);
	}
	return 0;
}


 

【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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