I'm sure there is a better solution, but this is the only thing I came up with that worked (for http://tour.golang.org/#69)

GoEquivalentBinaryTrees.go

package main

import "code.google.com/p/go-tour/tree"
import "fmt"
import "sort"
    

// Walk walks the tree t sending all values
// from the tree to the channel ch.
func Walk(t *tree.Tree, ch chan int) {
  WalkInternal(t, ch)
  close(ch)
}

func WalkInternal(t *tree.Tree, ch chan int) {
  if t.Left != nil {
    //fmt.Println(t.Value, "going left")
    WalkInternal(t.Left, ch)
  }
  if t.Right != nil {
    //fmt.Println(t.Value, "going right")
    WalkInternal(t.Right, ch)
  }
  //fmt.Println(t.Value, "visiting")
  ch <- t.Value
  //fmt.Println(t.Value, "going up")
    
  
}


// Same determines whether the trees
// t1 and t2 contain the same values.
func Same(t1, t2 *tree.Tree) bool {
  ch1 := make(chan int)
  ch2 := make(chan int)
  go Walk(t1, ch1)
  go Walk(t2, ch2)
  xs := make([]int, 0)
  ys := make([]int, 0)    
    
  for i := 0; i < 10; i++ {
      x:= <-ch1
      y:= <-ch2
      //fmt.Println(x, y)
      xs = append(xs, x)
      ys = append(ys, y)
  }
  sort.Ints(xs)
  sort.Ints(ys)
  
  return isEq(xs, ys)
}

func main() {
  //ch := make(chan int)
  //go Walk(tree.New(1), ch)
  //for i := range ch {
  //   fmt.Println(i)
  //}
  //  for i := 0; i < 10; i++ {
  //      fmt.Println(<-ch)
  //  }
    
  fmt.Println(Same(tree.New(1), tree.New(1)))
  fmt.Println(Same(tree.New(1), tree.New(2)))
}

func isEq(a, b []int) bool {
  for i, v := range a {
    if v != b[i] {
      return false
    }
  }
  return true
}