Exercise: Equivalent Binary Trees

tree.go

package main

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

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

func WalkRecursive(t *tree.Tree, ch chan int) {
    if t != nil {
        WalkRecursive(t.Left, ch)
        ch <- t.Value
        WalkRecursive(t.Right, ch)
    }
}

// Same determines whether the trees
// t1 and t2 contain the same values.
func Same(t1, t2 *tree.Tree) bool {
    ch1, ch2 := make(chan int), make(chan int)
    go Walk(t1, ch1)
    go Walk(t2, ch2)
    for {
        n1, ok1 := <- ch1
        n2, ok2 := <- ch2
        if ok1 != ok2 || n1 != n2 {
        	return false
        }
        if !ok1 {
        	break;
        }
    }
    return true
    
    
}

func main() {
    ch := make(chan int)
    go Walk(tree.New(1), ch)
    fmt.Println(Same(tree.New(1), tree.New(2)))
    fmt.Println(Same(tree.New(1), tree.New(1)))
    fmt.Println(Same(tree.New(2), tree.New(1)))
    
}

The same function must check whether t1 and t2 have the same values in any order. The implementation returns instead, whether the order is of the elements the same in both parameters.

The requirement specifically states:

constructs a randomly-structured (but always sorted) binary tree

So, the solutions provided are all correct; they rely on the tree always being sorted. Not much use for a binary tree if it wouldn't be, anyhow

sorted and balanced are different things. The functions written above do not work.
For example, the tree.New(1) function can return the following trees:

They are both sorted, but the structure is different. The functions written above return false for these two trees, although they should be true