Binary search tree implementation in go language

This is a creation in Article, where the information may have evolved or changed.

Using go to achieve the next two-fork search tree, or spent a lot of time, in the implementation of the use of a one-way list, this is to realize the two-way list in the realization of the advantages

Package Datastructureimport ("Container/list", "FMT") type bstree struct {root *node}type Node struct {left *noderight *node Value int}//newbstree Create tree func newbstree () *bstree {return &bstree{}}func (t *bstree) Insert (value int) {var parent *n Odez: = &node{value:value}x: = t.rootfor x! = Nil {parent = Xif Z.value < x.value {x = x.left} else {x = x.right}} if parent = = Nil {//the tree is empty t.root = Z} else if z.value < parent.value {parent.left = Z} else {parent.right = Z}}func (T * Bstree) Search (x int) *node {node: = t.rootfor Node! = Nil {if Node.value = = x {return Node} else if x < Node.value {No de = node.left} else {node = Node.right}}return nil}func (t *bstree) Delete (x int) bool {var parent *nodenode: = T.ROOTISF IND: = falsefor Node! = Nil {if Node.value = = x {isfind = truebreak} else if x < node.value {parent = Nodenode = Node.l EFT} else {parent = Nodenode = Node.right}}if Isfind = = False {return false}//case one: node is leaf node if node.left = Nil && n Ode.right = = Nil {if parent = nil {t.root = nil} else {if Parent.left = = Node {parent.left = nil} else {parent.right = Nil}}return true}//Situation II: Left child The side is empty or the right child is empty if Node.left = = Nil | | Node.right = = Nil {if parent = = Nil {if Node.left = nil {t.root = node.right} else {t.root = Node.left}} else {if parent. left = = Node {if Node.left = nil {parent.left = node.right} else {parent.left = Node.left}} else {if Node.left = nil {PA    Rent.right = node.right} else {parent.right = Node.left}}} return True}//condition three: Two children are not empty re: = Node.left Re_parent: = node for re.right! = Nil {//Find the parent node of the precursor node and the precursor node re_parent = re re = re.right} n    Ode.value = re.value if node = = re_parent {node.left = re.left} else {re_parent.right = Re.left } return true}//printtree1 recursive struct func (t *node) PrintTree1 () {if t.left! = nil {t.left.printtree1 ()}fmt. Print (T.value, "") if t.right! = nil {t.right.printtree1 ()}}//printtree2 non-recursive struct func (t *node) PrintTree2 () {stack: = list. New () stack. Pushback (t) for {node: = stack. Back () if node = = Nil {return}stack. Remove (node) v, _: = node. Value. (*node) fmt. Println (v.value) if v.left! = nil {stack. Pushback (v.left)}if v.right! = nil {stack. Pushback (V.right)}}}