Binary Tree Level Order Traversal Solutions in Go

package leetcode

import (
	"github.com/halfrost/LeetCode-Go/structures"
)

// TreeNode define
type TreeNode = structures.TreeNode

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */

// 解法一 BFS
func levelOrder(root *TreeNode) [][]int {
	if root == nil {
		return [][]int{}
	}
	queue := []*TreeNode{}
	queue = append(queue, root)
	curNum, nextLevelNum, res, tmp := 1, 0, [][]int{}, []int{}
	for len(queue) != 0 {
		if curNum > 0 {
			node := queue[0]
			if node.Left != nil {
				queue = append(queue, node.Left)
				nextLevelNum++
			}
			if node.Right != nil {
				queue = append(queue, node.Right)
				nextLevelNum++
			}
			curNum--
			tmp = append(tmp, node.Val)
			queue = queue[1:]
		}
		if curNum == 0 {
			res = append(res, tmp)
			curNum = nextLevelNum
			nextLevelNum = 0
			tmp = []int{}
		}
	}
	return res
}

// 解法二 DFS
func levelOrder1(root *TreeNode) [][]int {
	levels := [][]int{}
	dfsLevel(root, -1, &levels)
	return levels
}

func dfsLevel(node *TreeNode, level int, res *[][]int) {
	if node == nil {
		return
	}
	currLevel := level + 1
	for len(*res) <= currLevel {
		*res = append(*res, []int{})
	}
	(*res)[currLevel] = append((*res)[currLevel], node.Val)
	dfsLevel(node.Left, currLevel, res)
	dfsLevel(node.Right, currLevel, res)
}