Published Feb 25, 2022
[
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).
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
Input: root = [1]
Output: [[1]]
Input: root = []
Output: []
- The number of nodes in the tree is in the range [0, 2000].
- -1000 <= Node.val <= 1000
/**
* 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[][] {
let res: number[][] = [];
let queue: TreeNode[] = [];
let count: number = 1;
queue.push(root);
while(queue.length !== 0){
let temp: number[] = [];
const SIZE: number = queue.length
for(let i = 0; i < SIZE; i++){
let front: TreeNode = queue.shift();
if(front !== null){
temp.push(front.val)
queue.push(front.left);
queue.push(front.right);
}
}
count *= 2;
if(temp.length !== 0) res.push(temp);
}
return res;
};