LeetCode-in-Go
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
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func averageOfLevels(root *TreeNode) []float64 {
levelSums := make(map[int][]float64)
helper(root, levelSums, 0)
result := make([]float64, len(levelSums))
for level, pair := range levelSums {
result[level] = pair[1] / pair[0]
}
return result
}
func helper(root *TreeNode, levelSums map[int][]float64, level int) {
if root == nil {
return
}
if _, exists := levelSums[level]; !exists {
levelSums[level] = []float64{0, 0}
}
pair := levelSums[level]
pair[0]++ // count
pair[1] += float64(root.Val) // sum
levelSums[level] = pair
helper(root.Left, levelSums, level+1)
helper(root.Right, levelSums, level+1)
}