1. 二叉树的定义
二叉树需满足的条件
① 本身是有序树
② 树中包含的各个节点的长度不能超过2，即只能是0、1或者2
2. 前序遍历
前序遍历二叉树的顺序：根——》左——》右
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package mainimport "fmt"//定义结构体type Student struct { Name string Age int Score float32 left *Student //左子树指针 right *Student //右子树指针}//二叉树定义func main() { //根节点 var root Student root.Name = "root" root.Age = 18 root.Score = 88 //一级左子树 var left1 Student left1.Name = "left1" left1.Age = 20 left1.Score = 80 root.left = &left1 //一级右子树 var right1 Student right1.Name = "right1" right1.Age = 22 right1.Score = 100 root.right = &right1 //二级左子树 var left2 Student left2.Name = "left2" left2.Age = 25 left2.Score = 90 left1.left = &left2 //调用遍历函数 Req(&root)}//递归算法遍历整个二叉树func Req(tmp *Student) { for tmp == nil { return } fmt.Println(tmp) //遍历左子树 Req(tmp.left) //遍历右子树 Req(tmp.right)} |
输出结果如下
&{root 18 88 0xc0000c0480 0xc0000c04b0}
&{left1 20 80 0xc0000c04e0 <nil>}
&{left2 25 90 <nil> <nil>}
&{right1 22 100 <nil> <nil>}
3. 中序遍历
中序遍历：左——》根——》右
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package mainimport "fmt"//定义结构体type Student struct { Name string Age int Score float32 left *Student //左子树指针 right *Student //右子树指针}//二叉树定义func main() { //根节点 var root Student root.Name = "root" root.Age = 18 root.Score = 88 //一级左子树 var left1 Student left1.Name = "left1" left1.Age = 20 left1.Score = 80 root.left = &left1 //一级右子树 var right1 Student right1.Name = "right1" right1.Age = 22 right1.Score = 100 root.right = &right1 //二级左子树 var left2 Student left2.Name = "left2" left2.Age = 25 left2.Score = 90 left1.left = &left2 //调用遍历函数 Req(&root)}//递归算法遍历整个二叉树func Req(tmp *Student) { for tmp == nil { return } //遍历左子树 Req(tmp.left) //输出root节点 fmt.Println(tmp) //遍历右子树 Req(tmp.right)} |
输出结果如下
&{left2 25 90 <nil> <nil>}
&{left1 20 80 0xc000114510 <nil>}
&{root 18 88 0xc0001144b0 0xc0001144e0}
&{right1 22 100 <nil> <nil>}
4. 后序遍历
后序遍历：左——》右——》根
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package mainimport "fmt"//定义结构体type Student struct { Name string Age int Score float32 left *Student //左子树指针 right *Student //右子树指针}//二叉树定义func main() { //根节点 var root Student root.Name = "root" root.Age = 18 root.Score = 88 //一级左子树 var left1 Student left1.Name = "left1" left1.Age = 20 left1.Score = 80 root.left = &left1 //一级右子树 var right1 Student right1.Name = "right1" right1.Age = 22 right1.Score = 100 root.right = &right1 //二级左子树 var left2 Student left2.Name = "left2" left2.Age = 25 left2.Score = 90 left1.left = &left2 //调用遍历函数 Req(&root)}//递归算法遍历整个二叉树func Req(tmp *Student) { for tmp == nil { return } //遍历左子树 Req(tmp.left) //遍历右子树 Req(tmp.right) //输出root节点 fmt.Println(tmp)} |
输出结果如下
&{left2 25 90 <nil> <nil>}
&{left1 20 80 0xc0000c04e0 <nil>}
&{right1 22 100 <nil> <nil>}
&{root 18 88 0xc0000c0480 0xc0000c04b0}
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原文链接：https://www.cnblogs.com/lvrui/p/16163126.html
