If /g/ is so smart answer this >Write a recursive method

>>58437277
/g/ - homework

Do your own homework, Raj.

var compTree = function(tree1, tree2) {
  var same = true;

  if (tree1.left !== tree2.left) {
    same = false;
  }
  if (tree1.right !== tree2.right) {
    same = false;
  }

  return same && compTree(tree1.left, tree2.left) && compTree(tree1.right, tree2.right)
}

btw this will not work

Ok

struct tree {
    int payload;
    struct tree *left;
    struct tree *right;
};

bool compTree (struct tree *t1, struct tree *t2)
{
    if ((t1 == NULL) && (t2 == NULL))
        return true;
    if ((t1 == NULL) ^ (t2 == NULL))
        return false;
    return (t1->payload == t2->payload) && compTree(t1->left, t2->left) && compTree(t1->right, t2->right);
}

bool
compTree(Tree t1, Tree t2)
{
    return t1 == t2;
}

If they have the same root and the same pointers to their left and right subtrees, then they are the same.

Now back to the homework board:
>>>/hm/

void compTree(t1, t2) {

    // int128 so we can compute BIG numbers 4u
    int128 t1, t2;

    // this is where the magic happens

while (i=t1/t2) do {
    recurse(int128.to.const_char*(t1), int128.to.const_char*(t2));
    }

return 0;
}

;; assume we have tree?, tree-left, and tree-right to
;; check if is tree, access left child, and access right child, 
(define (compTree t1 t2)
 (if (and (tree? t1) (tree? t2))
     (and (compTree (tree-left t1) (tree-left t2))
          (compTree (tree-right t1) (tree-right t2)))
     (eq? t1 t2)))

Doing this efficiently (without blowing the stack) is left as an exercise to the reader.