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
}