LeetCode in TypeScript
637. Average of Levels in Binary Tree
Easy
Given the root of a binary tree, return the average value of the nodes on each level in the form of an array. Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [3.00000,14.50000,11.00000] Explanation: The average value of nodes on level 0 is 3, on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11].
Example 2:
Input: root = [3,9,20,15,7]
Output: [3.00000,14.50000,11.00000]
Constraints:
- The number of nodes in the tree is in the range
[1, 104]. -231 <= Node.val <= 231 - 1
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 averageOfLevels(root: TreeNode | null): number[] {
if (!root) {
return []
}
let res = []
let queue = [root]
while (queue.length) {
let len = queue.length
let sum = 0
for (let i = 0; i < len; i++) {
let node = queue.shift()
sum += node.val
if (node.left) {
queue.push(node.left)
}
if (node.right) {
queue.push(node.right)
}
}
res.push(sum / len)
}
return res
}
export { averageOfLevels }