simple binary tree implementation and printer for those pesky interview problems

tree.go

package main

import (
	"fmt"
	"strings"
)

type BinaryTree struct {
	Node  int // or whatever
	Left  *BinaryTree
	Right *BinaryTree
}

func (t BinaryTree) String() string {
	return t.pretty(0)
}

func (t BinaryTree) pretty(depth int) string {
	str := leftPad(fmt.Sprintf("<tree value='%v'>", t.Node), depth)
	if t.Left != nil {
		str += "\n" + t.Left.pretty(depth+1)
	}
	if t.Right != nil {
		str += "\n" + t.Right.pretty(depth+1)
	}
	if t.Right == nil && t.Left == nil {
		str += "</tree>"
	} else {
		str += "\n" + leftPad("</tree>", depth)
	}
	return str
}

type Tree struct {
	Node     int // or whatever
	Children []*Tree
}

func (t Tree) String() string {
	return t.pretty(0)
}

func (t Tree) pretty(depth int) string {
	str := leftPad(fmt.Sprintf("<tree value='%v'>", t.Node), depth)
	hasChildren := false
	for _, child := range t.Children {
		hasChildren = true
		str += "\n" + child.pretty(depth+1)
	}
	if !hasChildren {
		str += "</tree>"
	} else {
		str += "\n" + leftPad("</tree>", depth)
	}
	return str
}

func leftPad(s string, depth int) string {
	return strings.Repeat(" ", depth) + s
}

func main() {
	btree := &BinaryTree{Node: 0}
	btree.Left = &BinaryTree{Node: -1}
	btree.Right = &BinaryTree{Node: 1, Right: &BinaryTree{Node: 2}}
	fmt.Println(btree)

	tree := &Tree{
		Node: 0,
		Children: []*Tree{
			&Tree{Node: -1},
			&Tree{Node: 1, Children: []*Tree{
				&Tree{Node: 2},
			}},
		},
	}
	fmt.Println(tree)
}

Output:

<tree value='0'>
 <tree value='-1'></tree>
 <tree value='1'>
  <tree value='2'></tree>
 </tree>
</tree>
<tree value='0'>
 <tree value='-1'></tree>
 <tree value='1'>
  <tree value='2'></tree>
 </tree>
</tree>