Given a binary tree, design an algorithm which creates a linked list of all the nodes at each depth (e.g., if you have a tree with depth D, you'll have D linked lists). Return a array containing all the linked lists.
Example:
Input: [1,2,3,4,5,null,7,8]
1
/ \
2 3
/ \ \
4 5 7
/
8
Output: [[1],[2,3],[4,5,7],[8]]
Level order traversal.
# 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]:
q = [tree]
ans = []
while q:
n = len(q)
head = ListNode(-1)
tail = head
for i in range(n):
front = q.pop(0)
node = ListNode(front.val)
tail.next = node
tail = node
if front.left:
q.append(front.left)
if front.right:
q.append(front.right)
ans.append(head.next)
return ans/**
* 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) {
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(tree);
List<ListNode> ans = new ArrayList<>();
while (!queue.isEmpty()) {
int n = queue.size();
ListNode head = new ListNode(-1);
ListNode tail = head;
for (int i = 0; i < n; i++) {
TreeNode front = queue.poll();
ListNode node = new ListNode(front.val);
tail.next = node;
tail = node;
if (front.left != null) {
queue.offer(front.left);
}
if (front.right != null) {
queue.offer(front.right);
}
}
ans.add(head.next);
}
return ans.toArray(new ListNode[0]);
}
}/**
* 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;
if (tree == nullptr) {
return ans;
}
queue<TreeNode*> q;
q.push(tree);
while (!q.empty()) {
int n = q.size();
ListNode* head = new ListNode(-1);
ListNode* tail = head;
for (int i = 0; i < n; ++i) {
TreeNode* front = q.front();
q.pop();
ListNode* node = new ListNode(front->val);
tail->next = node;
tail = node;
if (front->left != nullptr) {
q.push(front->left);
}
if (front->right != nullptr) {
q.push(front->right);
}
}
ans.push_back(head->next);
}
return ans;
}
};/**
* 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 {
queue := make([]*TreeNode, 0)
queue = append(queue, tree)
ans := make([]*ListNode, 0)
for len(queue) > 0 {
n := len(queue)
head := new(ListNode)
tail := head
for i := 0; i < n; i++ {
front := queue[0]
queue = queue[1:]
node := &ListNode{Val: front.Val}
tail.Next = node
tail = node
if front.Left != nil {
queue = append(queue, front.Left)
}
if front.Right != nil {
queue = append(queue, front.Right)
}
}
ans = append(ans, head.Next)
}
return ans
}/**
* 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 res = [];
if (tree == null) {
return res;
}
const queue = [tree];
while (queue.length !== 0) {
const n = queue.length;
const dummy = new ListNode();
let cur = dummy;
for (let i = 0; i < n; i++) {
const { val, left, right } = queue.shift();
left && queue.push(left);
right && queue.push(right);
cur.next = new ListNode(val);
cur = cur.next;
}
res.push(dummy.next);
}
return res;
}// 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;
use std::collections::VecDeque;
impl Solution {
pub fn list_of_depth(tree: Option<Rc<RefCell<TreeNode>>>) -> Vec<Option<Box<ListNode>>> {
let mut res = vec![];
if tree.is_none() {
return res;
}
let mut q = VecDeque::new();
q.push_back(tree);
while !q.is_empty() {
let n = q.len();
let mut demmy = Some(Box::new(ListNode::new(0)));
let mut cur = &mut demmy;
for _ in 0..n {
if let Some(node) = q.pop_front().unwrap() {
let mut node = node.borrow_mut();
if node.left.is_some() {
q.push_back(node.left.take());
}
if node.right.is_some() {
q.push_back(node.right.take());
}
cur.as_mut().unwrap().next = Some(Box::new(ListNode::new(node.val)));
cur = &mut cur.as_mut().unwrap().next;
}
}
res.push(demmy.as_mut().unwrap().next.take());
}
res
}
}