2021-12-09：二叉树展开为链表。 给你二叉树的根结点 root ，请你将它展开为一个单链表： 展开后的单链表应该同样使用

package main

import "fmt"

func main() {
    head := NewTreeNode(1)
    head.left = NewTreeNode(2)
    head.right = NewTreeNode(3)
    head = convert(head)
    for head != nil {
        fmt.Println(head.val)
        head = head.right
    }
}

type TreeNode struct {
    val   int
    left  *TreeNode
    right *TreeNode
}

func NewTreeNode(value int) *TreeNode {
    ret := &TreeNode{}
    ret.val = value
    return ret
}

func convert(head *TreeNode) *TreeNode {
    if head == nil {
        return nil
    }
    return process(head).head
}

type Info struct {
    head *TreeNode
    tail *TreeNode
}

func NewInfo(h, t *TreeNode) *Info {
    res := &Info{}
    res.head = h
    res.tail = t
    return res
}

func process(x *TreeNode) *Info {
    if x == nil {
        return nil
    }
    leftInfo := process(x.left)
    rightInfo := process(x.right)
    // 2...4 5 6...13
    if leftInfo != nil {
        leftInfo.tail.right = x
        x.left = leftInfo.tail
    }
    if rightInfo != nil {
        x.right = rightInfo.head
        rightInfo.head.left = x
    }
    var head *TreeNode
    if leftInfo != nil {
        head = leftInfo.head
    } else {
        head = x
    }
    var tail *TreeNode
    if rightInfo != nil {
        tail = rightInfo.tail
    } else {
        tail = x
    }
    return NewInfo(head, tail)
}