算法leetcode面试题 04.03. 特定深度节点链表（多语言实现）
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面试题 04.03. 特定深度节点链表：
给定一棵二叉树，设计一个算法，创建含有某一深度上所有节点的链表（比如，若一棵树的深度为 D，则会创建出 D 个链表）。返回一个包含所有深度的链表的数组。
样例 1：
输入：
[1,2,3,4,5,null,7,8]
1
/ \\
2 3
/ \\ \\
4 5 7
/
8
输出：
[[1],[2,3],[4,5,7],[8]]
分析
- 面对这道算法题目，二当家的陷入了沉思。
- 如果对二叉树和链表比较熟悉，就会明白其实就是二叉树的层序遍历，每一层组合成一条链表。
题解
rust
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
//
//
// impl TreeNode
// #[inline]
// pub fn new(val: i32) -> Self
// TreeNode
// val,
// left: None,
// right: None
//
//
//
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode
// pub val: i32,
// pub next: Option<Box<ListNode>>
//
//
// impl ListNode
// #[inline]
// fn new(val: i32) -> Self
// ListNode
// next: None,
// val
//
//
//
use std::rc::Rc;
use std::cell::RefCell;
impl Solution
pub fn list_of_depth(tree: Option<Rc<RefCell<TreeNode>>>) -> Vec<Option<Box<ListNode>>>
let mut ans: Vec<Option<Box<ListNode>>> = Vec::new();
let mut queue = std::collections::VecDeque::new();
queue.push_back(tree);
while !queue.is_empty()
let mut head = Some(Box::new(ListNode::new(0)));
let mut tail = head.as_mut();
let size = queue.len();
for _ in 0..size
let node = queue.pop_front().unwrap().unwrap();
let mut node = node.borrow_mut();
if node.left.is_some()
queue.push_back(node.left.take());
if node.right.is_some()
queue.push_back(node.right.take());
tail.as_mut().unwrap().next = Some(Box::new(ListNode::new(node.val)));
tail = tail.unwrap().next.as_mut();
ans.push(head.as_mut().unwrap().next.take());
ans
go
/**
* Definition for a binary tree node.
* type TreeNode struct
* Val int
* Left *TreeNode
* Right *TreeNode
*
*/
/**
* Definition for singly-linked list.
* type ListNode struct
* Val int
* Next *ListNode
*
*/
func listOfDepth(tree *TreeNode) []*ListNode
var ans []*ListNode
queue := []*TreeNodetree
for len(queue) > 0
head := &ListNode
tail := head
size := len(queue)
for i := 0; i < size; i++
node := queue[i]
if node.Left != nil
queue = append(queue, node.Left)
if node.Right != nil
queue = append(queue, node.Right)
tail.Next = &ListNodeVal: node.Val
tail = tail.Next
ans = append(ans, head.Next)
queue = queue[size:]
return ans
typescript
/**
* Definition for a binary tree node.
* class TreeNode
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null)
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
*
*
*/
/**
* Definition for singly-linked list.
* class ListNode
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null)
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
*
*
*/
function listOfDepth(tree: TreeNode | null): Array<ListNode | null>
const ans = [];
const queue = [tree];
while (queue.length > 0)
const head = new ListNode();
let tail = head;
const size = queue.length;
for (let i = 0; i < size; ++i)
const val, left, right = queue.shift();
left && queue.push(left);
right && queue.push(right);
tail.next = new ListNode(val);
tail = tail.next;
ans.push(head.next);
return ans;
;
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def listOfDepth(self, tree: TreeNode) -> List[ListNode]:
ans = []
q = collections.deque()
q.append(tree)
while len(q) > 0:
head = ListNode()
tail = head
size = len(q)
for _ in range(size):
node = q.popleft()
node.left and q.append(node.left)
node.right and q.append(node.right)
tail.next = ListNode(node.val)
tail = tail.next
ans.append(head.next)
return ans
c
/**
* Definition for a binary tree node.
* struct TreeNode
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* ;
*/
/**
* Definition for singly-linked list.
* struct ListNode
* int val;
* struct ListNode *next;
* ;
*/
int getDepth(struct TreeNode* tree)
if (!tree)
return 0;
int leftDepth = getDepth(tree->left);
int rightDepth = getDepth(tree->right);
return fmax(leftDepth, rightDepth) + 1;
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
struct ListNode** listOfDepth(struct TreeNode* tree, int* returnSize)
int depth = getDepth(tree);
struct ListNode **ans = malloc(depth * sizeof(struct ListNode *));
*returnSize = 0;
struct TreeNode *queue[(int) pow(2, depth) - 1];
queue[0] = tree;
int start = 0;
int end = 1;
while (start < end)
struct ListNode head = ;
struct ListNode *tail = &head;
int curEnd = end;
while (start < curEnd)
struct TreeNode *node = queue[start++];
if (node->left)
queue[end++] = node->left;
if (node->right)
queue[end++] = node->right;
tail->next = malloc(sizeof(struct ListNode));
tail->next->val = node->val;
tail->next->next = NULL;
tail = tail->next;
ans[(*returnSize)++] = head.next;
return ans;
c++
/**
* Definition for a binary tree node.
* struct TreeNode
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL)
* ;
*/
/**
* Definition for singly-linked list.
* struct ListNode
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL)
* ;
*/
class Solution
public:
vector<ListNode*> listOfDepth(TreeNode* tree)
vector<ListNode *> ans;
queue<TreeNode *> q;
q.push(tree);
while (q.size() > 0)
ListNode head = ListNode(0);
ListNode *tail = &head;
int size = q.size();
for (int i = 0; i < size; ++i)
TreeNode *node = q.front();
q.pop();
if (node->left != NULL)
q.push(node->left);
if (node->right != NULL)
q.push(node->right);
tail->next = new ListNode(node->val);
tail = tail->next;
ans.emplace_back(head.next);
return ans;
;
java
/**
* 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
public ListNode[] listOfDepth(TreeNode tree)
List<ListNode> list = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
queue.add(tree);
while (!queue.isEmpty())
ListNode head = new ListNode();
ListNode tail = head;
int size = queue.size();
for (int i = 0; i < size; ++i)
TreeNode node = queue.poll();
if (node.left != null)
queue.add(node.left);
if (node.right != null)
queue.add(node.right);
tail.next = new ListNode(node.val);
tail = tail.next;
list.add(head.next);
ListNode[] ans = new ListNode[list.size()];
list.toArray(ans);
return ans;
原题传送门：https://leetcode.cn/problems/list-of-depth-lcci/submissions/
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