GO函数式编程（二叉树应用）

先看一下通常的二叉树写法，上代码：

package main

import (
	"fmt"
)

type Node struct {
	LeftNode *Node
	RightNode *Node
	Value int
}

func buildTree() Node {
	var tree Node
	tree.Value = 0

	tree.LeftNode = new(Node)
	tree.LeftNode.Value = 1

	tree.LeftNode.LeftNode = new(Node)
	tree.LeftNode.LeftNode.Value = 3

	tree.LeftNode.LeftNode.LeftNode = new(Node)
	tree.LeftNode.LeftNode.LeftNode.Value = 5

	tree.LeftNode.LeftNode.RightNode = new(Node)
	tree.LeftNode.LeftNode.RightNode.Value = 6

	tree.RightNode = new(Node)
	tree.RightNode.Value = 2

	tree.RightNode.LeftNode = new(Node)
	tree.RightNode.LeftNode.Value = 4

	tree.RightNode.LeftNode.RightNode = new(Node)
	tree.RightNode.LeftNode.RightNode.Value = 7

	return tree
}

func (node Node)do()  {
	fmt.Println(node.Value)
	if node.LeftNode != nil {
		node.LeftNode.do()
	}
	if node.RightNode != nil{
		node.RightNode.do()
	}
}

func main() {
	tree := buildTree()
	tree.do()
}

函数式编程的写法：

node.go

package node

type Node struct {
	LeftNode *Node
	RightNode *Node
	Value int
}

func (node Node)Do(f func(n *Node))  {
	f(&node)
	if node.LeftNode != nil {
		node.LeftNode.Do(f)
	}
	if node.RightNode != nil{
		node.RightNode.Do(f)
	}
}

main.go

	package main
	
	import (
		"fmt"
		"go_study/005_function/treenode/node"
		"strconv"
	)
	
	
	func buildTree() node.Node {
		var tree node.Node
		tree.Value = 0
	
		tree.LeftNode = new(node.Node)
		tree.LeftNode.Value = 1
	
		tree.LeftNode.LeftNode = new(node.Node)
		tree.LeftNode.LeftNode.Value = 3
	
		tree.LeftNode.LeftNode.LeftNode = new(node.Node)
		tree.LeftNode.LeftNode.LeftNode.Value = 5
	
		tree.LeftNode.LeftNode.RightNode = new(node.Node)
		tree.LeftNode.LeftNode.RightNode.Value = 6
	
		tree.RightNode = new(node.Node)
		tree.RightNode.Value = 2
	
		tree.RightNode.LeftNode = new(node.Node)
		tree.RightNode.LeftNode.Value = 4
	
		tree.RightNode.LeftNode.RightNode = new(node.Node)
		tree.RightNode.LeftNode.RightNode.Value = 7
	
		return tree
	}
	
	func main() {
		tree := buildTree()
		count := 0
		tree.Do(func(n *node.Node) {
			count++
			fmt.Println("count : "+strconv.Itoa(count))
		})
	
		tree.Do(func(n *node.Node) {
	
			fmt.Println("value : "+strconv.Itoa(n.Value))
		})
	}

可以看出函数式编程比较灵活一些，可以灵活的做你想要做到的事情