4 4 Given a binary search tree, design an algorithm which creates a linked list of all the nodes at

本文详细阐述了如何通过遍历给定的二叉搜索树,为每一层生成独立的链表。具体操作包括初始化链表、递归地遍历树结构并收集各层级的节点,最终实现每层节点的有序排列。实例代码展示了从根节点开始,通过深度优先搜索策略构建层次化节点列表的过程。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

4 4 Given a binary search tree, design an algorithm which creates a linked list of all thenodes at each depth (eg, if you have a tree with depth D, you’ll have D linked lists) 


package Tree;

import java.util.ArrayList;
import java.util.LinkedList;

/**
 * @Title: Tree_D_dep_Dlist.java
 * @Package Tree
 * @Description: TODO
 * @author nutc
 * @date 2013-8-21 下午2:25:37
 * @version V1.0
 */
public class Tree_D_dep_Dlist {

	public static void main(String args[]) {
		Node node1 = new Node(1);
		Node node2 = new Node(2);
		Node node3 = new Node(3);
		Node node4 = new Node(4);
		Node node5 = new Node(5);
		Node node6 = new Node(6);

		node3.addLeft(node1);
		 node3.addRight(node5);
		 node5.addLeft(node4);
		node1.addRight(node2);
		 node5.addRight(node6);

		Node node9 = new Node(9);
		node1.addLeft(node9);

		node3.display();
		ArrayList<LinkedList<Node>> map = find(node3);
		for (int i = 0; i < map.size(); i++) {
			System.out.println("当前层数" + i);
			for (int j = 0; j < map.get(i).size(); j++) {
				System.out.print(map.get(i).get(j).value + " ");
			}
			System.out.println();
		}
	}

	public static ArrayList<LinkedList<Node>> find(Node root) {
		if (root == null)
			return null;

		ArrayList<LinkedList<Node>> map = new ArrayList<LinkedList<Node>>();
		int level = 0;
		LinkedList<Node> nowlist = new LinkedList<Node>();
		nowlist.push(root);
		map.add(nowlist);

		while (map.size()>level ) {  //这里原来用map.get(level)是错误滴,超出下表了
			nowlist = map.get(level);
			LinkedList<Node> templist = new LinkedList<Node>();
			for (int i = 0; i < nowlist.size(); i++) {
				if (nowlist.get(i).left != null)
					templist.add(nowlist.get(i).left);
				if (nowlist.get(i).right != null)
					templist.add(nowlist.get(i).right);
			}
			level++;   //这个地方,忘记++了。。不过这些是因为使用的结构改动过造成的。。不过! 还是要仔细检查!
			if (templist.size() > 0)
				map.add(templist);
		}

		return map;
	}
}


【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.
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值