二叉树:如何用golang将二叉树展开为链表-新版实现逻辑

接上一篇，用迭代实现了将链表合并成一个链表，我感觉看着那代码让我发晕，所以我参考B占一个博主的思路，换了一个实现方式，代码跟简洁，思路跟清晰，如下是我的完整代码：

package main

import (
    "fmt"
    "testing"
)

type (
	TreeNd struct {
		val   int
		left  *TreeNd
		right *TreeNd
	}
)

func createNode(value int) *TreeNd {
    return &TreeNd{
       val:   value,
       left:  nil,
       right: nil,
    }
}

func flatten(root *TreeNd) *TreeNd {
    if root == nil {
       return nil
    }

    cur := root
    for cur != nil {
       if cur.left != nil {
          leftBottomNode := cur.left
          for leftBottomNode.right != nil {
             leftBottomNode = leftBottomNode.right
          }
          leftBottomNode.right = cur.right
          cur.right = cur.left
          cur.left = nil
       }
       cur = cur.right
    }

    return root
}

func convert2Arr(node *TreeNd) []int {
	if node == nil {
		return []int{}
	}
	result := []int{node.val}
	result = append(result, convert2Arr(node.left)...)
	result = append(result, convert2Arr(node.right)...)
	return result
}

func TestDemo3(t *testing.T) {
    node1 := createNode(1)
    node2 := createNode(2)
    node3 := createNode(3)
    node4 := createNode(4)
    node5 := createNode(5)
    node6 := createNode(6)

    node1.left = node2
    node1.right = node5
    node2.left = node3
    node2.right = node4
    node5.right = node6

    root := flatten(node1)
    result := convert2Arr(root)
    fmt.Println("result => ", result)
}

大家看看是不是跟清晰了，跟简单