03. 特定深度节点链表

https://leetcode-cn.com/problems/list-of-depth-lcci/

Java

/*
 * @Author: Goog Tech
 * @Date: 2020-09-17 12:30:07
 * @LastEditTime: 2020-09-17 12:32:43
 * @Description: https://leetcode-cn.com/problems/list-of-depth-lcci/
 * @FilePath: \leetcode-googtech\面试题04\#03. 特定深度节点链表\Solution.java
 * @WebSite: https://algorithm.show/
 */

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */

class Solution {
    // BFS : 广度优先搜索,即层次遍历
    public ListNode[] listOfDepth(TreeNode tree) {
        // 判断头节点是否为空
        if(tree == null) return new ListNode[0];
        // 初始化辅助队列,并将根节点入队
        Queue<TreeNode> queue = new LinkedList<>(); 
        queue.add(tree);
        // 初始化链表集合
        List<ListNode> list = new ArrayList<>();
        // 初始化链表虚拟头节点
        ListNode dummyHead = new ListNode(-1); 
        while(!queue.isEmpty()) {
            // 初始化链表虚拟头节点的指针
            ListNode prev = dummyHead;
            // 获取当前队列中的节点数,即二叉树当前深度所有的节点数
            int size = queue.size();
            // 遍历队列中的节点
            for(int i = 0; i < size; i++) {
                // 获取队首节点
                TreeNode node  = queue.poll(); // or tree = queue.poll();
                // 将链表虚拟头节点的指针指向当前出队节点,即将其作为下一个节点
                prev.next = new ListNode(node.val);
                prev = prev.next;
                // 若当前出队节点的左右孩子不为空,则将其入队,进而对下个深度的节点进行遍历
                if(node.left != null) queue.offer(node.left);
                if(node.right != null) queue.offer(node.right);
            }
            // 将二叉树当前深度的节点所构成的链表添加到临时列表中
            list.add(dummyHead.next);
        }
        // 将临时列表中的元素存储到结果数组中,然后返回其结果
        ListNode[] result = new ListNode[list.size()];
        for(int i = 0; i < list.size(); i++) result[i] = list.get(i);
        return result;
    }
}

Python

'''
Author: Goog Tech
Date: 2020-09-17 12:30:16
LastEditTime: 2020-09-17 12:31:02
Description: https://leetcode-cn.com/problems/list-of-depth-lcci/
FilePath: \leetcode-googtech\面试题04\#03. 特定深度节点链表\Solution.py
WebSite: https://algorithm.show/
'''

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    # BFS : 广度优先搜索,即层次遍历
    def listOfDepth(self, tree):
        """
        :type tree: TreeNode
        :rtype: List[ListNode]
        """
        # 判断头节点是否为空
        if not tree: return []
        # 初始化模拟队列及结果列表
        queue, result = [tree], [] 
        # 判断队列是否为空
        while queue:
            # 初始化链表虚拟头节点及其指针
            dummyHead = ListNode(None)
            prev = dummyHead
            # 遍历辅助队列,其存储着二叉树当前深度的所有节点
            for i in range(len(queue)):
                # 获取队首节点
                node = queue.pop(0)
                # 将链表虚拟头节点的指针指向当前出队节点,即将其作为下一个节点
                prev.next = ListNode(node.val)
                prev = prev.next
                # 若当前出队节点的左右孩子不为空,则将其入队,进而对下个深度的节点进行遍历
                if node.left: queue.append(node.left)
                if node.right: queue.append(node.right)
            # 将二叉树当前深度的节点所构成的链表添加到结果列表中
            result.append(dummyHead.next)
        # 返回结果列表
        return result