首发于 语雀文档@blueju

    代码：

    1. /**
    2.  * Definition for a binary tree node.
    3.  * function TreeNode(val, left, right) {
    4.  *     this.val = (val===undefined ? 0 : val)
    5.  *     this.left = (left===undefined ? null : left)
    6.  *     this.right = (right===undefined ? null : right)
    7.  * }
    8.  */
    9. /**
    10.  * @param {TreeNode} root
    11.  * @return {number[]}
    12.  */
    13. var averageOfLevels = function (root) {
    14.     if (root === null) return []
    15.     /** 记录每层的和 */
    16.     const sum = []
    17.     /** 记录每层节点的数量 */
    18.     const count = []
    19.     dfs(root, 0, sum, count)
    20.     return sum.map((item, index) => {
    21.         return item / count[index]
    22.     })
    23. };
    25. function dfs(root, level, sum, count) {
    26.     if (root === null) return root
    27.     sum[level] = sum[level] === undefined ? root.val : sum[level] + root.val
    28.     count[level] = count[level] === undefined ? 1 : count[level] + 1
    29.     root.left && dfs(root.left, level + 1, sum, count)
    30.     root.right && dfs(root.right, level + 1, sum, count)
    31. }

    2021年11月2日再做一次

    1. /**
    2.  * Definition for a binary tree node.
    3.  * class TreeNode {
    4.  *     val: number
    5.  *     left: TreeNode | null
    6.  *     right: TreeNode | null
    7.  *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    8.  *         this.val = (val===undefined ? 0 : val)
    9.  *         this.left = (left===undefined ? null : left)
    10.  *         this.right = (right===undefined ? null : right)
    11.  *     }
    12.  * }
    13.  */
    15. function averageOfLevels(root: TreeNode | null): number[] {
    16.     if (root === null) return []
    17.     /* 每层树节点 */
    18.     let levelTreeNode = []
    19.     dfs(root, 1, levelTreeNode)
    20.     levelTreeNode.shift()
    21.     const result = levelTreeNode.map((item: number[]) => {
    22.         const sum = item.reduce((prev, curr) => prev + curr, 0)
    23.         return sum / item.length
    24.     })
    25.     return result
    26. };
    28. const dfs = (root: TreeNode, level: number, levelTreeNode) => {
    29.     if (Array.isArray(levelTreeNode[level])) {
    30.         levelTreeNode[level].push(root.val)
    31.     } else {
    32.         levelTreeNode[level] = []
    33.         levelTreeNode[level].push(root.val)
    34.     }
    35.     root.left && dfs(root.left, level + 1, levelTreeNode)
    36.     root.right && dfs(root.right, level + 1, levelTreeNode)
    37. }