429. N-ary Tree Level Order Traversal

Medium
​
Given an n-ary tree, return the level order traversal of its nodes' values.
Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).
Example 1:
Input: root = [1,null,3,2,4,null,5,6]
Output:
[[1],[3,2,4],[5,6]]
Example 2:
Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output:
[[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
Constraints:
  • The height of the n-ary tree is less than or equal to 1000
  • The total number of nodes is between [0, 104]

解題

以上面範例測資二為例,此程式碼放入的順序如下:
[]
[[1]]
[[1] [2]]
[[1] [2 3]]
[[1] [2 3] [6]]
[[1] [2 3] [6 7]]
[[1] [2 3] [6 7] [11]]
[[1] [2 3] [6 7] [11] [14]]
[[1] [2 3 4] [6 7] [11] [14]]
[[1] [2 3 4] [6 7 8] [11] [14]]
[[1] [2 3 4] [6 7 8] [11 12] [14]]
[[1] [2 3 4 5] [6 7 8] [11 12] [14]]
[[1] [2 3 4 5] [6 7 8 9] [11 12] [14]]
[[1] [2 3 4 5] [6 7 8 9] [11 12 13] [14]]
[[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
/**
* Definition for a Node.
* type Node struct {
* Val int
* Children []*Node
* }
*/
​
func levelOrder(root *Node) [][]int {
ans := make([][]int, 0)
​
var helper func(*Node, int)
helper = func(root *Node, index int) {
if root == nil {
return
}
​
if index == len(ans) {
ans = append(ans, []int{root.Val})
} else {
ans[index] = append(ans[index], root.Val)
}
​
for _, child := range root.Children {
helper(child, index + 1)
}
}
​
helper(root, 0)
​
return ans
}
​