Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.
Example 1:
Input:
3
/ \
9 20
/ \
15 7
Output: [3, 14.5, 11]
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]. Note: The range of node's value is in the range of 32-bit signed integer.
求二叉树每一层的平均数
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
type record struct {
val float64
num int
}
func averageOfLevels(root *TreeNode) []float64 {
res := make([]record, 0)
helper(root, &res, 1)
r := make([]float64, 0)
for _, v := range res {
r = append(r, v.val/float64(v.num))
}
return r
}
func helper(root *TreeNode, res *[]record, level int) {
if root == nil {
return
}
if len(*res) < level {
*res = append(*res, record{})
}
(*res)[level-1].val += float64(root.Val)
(*res)[level-1].num += 1
helper(root.Left, res, level+1)
helper(root.Right, res, level+1)
}