LeetCode in TypeScript
102. Binary Tree Level Order Traversal
Medium
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
Solution
import { TreeNode } from '../../com_github_leetcode/treenode'
/*
* 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[][] {
if (root == null) return []
let queue = [root]
let result = []
while (queue.length != 0) {
let subResult = []
let length = queue.length
while (length > 0) {
let node = queue.shift()
subResult.push(node.val)
if (node.left != null) queue.push(node.left)
if (node.right != null) queue.push(node.right)
length--
}
result.push(subResult)
}
return result
}
export { levelOrder }