0102.go

// Source: https://leetcode.com/problems/binary-tree-level-order-traversal
// Title: Binary Tree Level Order Traversal
// Difficulty: Medium
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).
//
// Example 1:
//
//   https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg
//
//   Input: root = [3,9,20,null,null,15,7]
//   Output: [[3],[9,20],[15,7]]
//
// Example 2:
//
//   Input: root = [1]
//   Output: [[1]]
//
// Example 3:
//
//   Input: root = []
//   Output: []
//
// Constraints:
//
//   The number of nodes in the tree is in the range [0, 2000].
//   -1000 <= Node.val <= 1000
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

package main

type TreeNode struct {
	Val   int
	Left  *TreeNode
	Right *TreeNode
}

func levelOrder(root *TreeNode) [][]int {
	res := [][]int{}

	_levelOrder(root, 0, &res)

	return res
}

func _levelOrder(root *TreeNode, depth int, res *[][]int) {
	if root == nil {
		return
	}

	if depth >= len(*res) {
		*res = append(*res, []int{})
	}

	(*res)[depth] = append((*res)[depth], root.Val)

	_levelOrder(root.Left, depth+1, res)
	_levelOrder(root.Right, depth+1, res)
}