# 102. Binary Tree Level Order Traversal

## Description

Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]


Example 2:

Input: root = [1]
Output: [[1]]


Example 3:

Input: root = []
Output: []


Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -1000 <= Node.val <= 1000

## Solutions

Solution 1: BFS

We can use the BFS method to solve this problem. First, enqueue the root node, then continuously perform the following operations until the queue is empty:

• Traverse all nodes in the current queue, store their values in a temporary array $t$, and then enqueue their child nodes.
• Store the temporary array $t$ in the answer array.

Finally, return the answer array.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
if (root == null) {
return ans;
}
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
while (!q.isEmpty()) {
List<Integer> t = new ArrayList<>();
for (int n = q.size(); n > 0; --n) {
TreeNode node = q.poll();
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
}
return ans;
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> ans;
if (!root) return ans;
queue<TreeNode*> q{ {root} };
while (!q.empty()) {
vector<int> t;
for (int n = q.size(); n; --n) {
auto node = q.front();
q.pop();
t.push_back(node->val);
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
ans.push_back(t);
}
return ans;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
ans = []
if root is None:
return ans
q = deque([root])
while q:
t = []
for _ in range(len(q)):
node = q.popleft()
t.append(node.val)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
ans.append(t)
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func levelOrder(root *TreeNode) (ans [][]int) {
if root == nil {
return
}
q := []*TreeNode{root}
for len(q) > 0 {
t := []int{}
for n := len(q); n > 0; n-- {
node := q[0]
q = q[1:]
t = append(t, node.Val)
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
ans = append(ans, t)
}
return
}

• /**
* 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)
*     }
* }
*/

function levelOrder(root: TreeNode | null): number[][] {
const res = [];
if (root == null) {
return res;
}
const queue = [root];
while (queue.length != 0) {
const n = queue.length;
res.push(
new Array(n).fill(null).map(() => {
const { val, left, right } = queue.shift();
left && queue.push(left);
right && queue.push(right);
return val;
}),
);
}
return res;
}


• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrder = function (root) {
let ans = [];
if (!root) {
return ans;
}
let q = [root];
while (q.length) {
let t = [];
for (let n = q.length; n; --n) {
const { val, left, right } = q.shift();
t.push(val);
left && q.push(left);
right && q.push(right);
}
ans.push(t);
}
return ans;
};


• // 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
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
let mut res = vec![];
if root.is_none() {
return res;
}
let mut queue: VecDeque<Option<Rc<RefCell<TreeNode>>>> = vec![root].into_iter().collect();
while !queue.is_empty() {
let n = queue.len();
res.push(
(0..n)
.into_iter()
.map(|_| {
let mut node = queue.pop_front().unwrap();
let mut node = node.as_mut().unwrap().borrow_mut();
if node.left.is_some() {
queue.push_back(node.left.take());
}
if node.right.is_some() {
queue.push_back(node.right.take());
}
node.val
})
.collect()
);
}
res
}
}