CareerCup

public boolean recurseSymmetry(Node<AnyType> left, Node<AnyType> right ){
if(left == null || right == null) return left==right;
else 
return left.value == right.value && recurseSymmetry(left.left, right.right) && recurseSymmetry(left.right, right.left);
}

Recursive checking
1. Each node contains 2 leaves, if none then all height must equal

main()
{
p=root->leftchild
q=root->rightchild
identical(&p,&q)
}
identical(struct btreenode*a,struct btreenode*b)
{
if((a==null&&b==null)
return true;
else if (a!=null&&b!=null)
{
return (a->data==b->data&& identical(a->left,b->right)&&identical(a->right,b->left);
}
else return false;
}

public boolean isSymmetric (BTNode current) {
		if (current == null) {
			return false;
		}

		if (current.left() != null && current.right() != null) {
			return isSymmetric(current.left()) && isSymmetric(current.right());
		} else {
			if (current.left() == null && current.right() == null) {
				return true;
			}
		}

		return false;
	}

Do a left->root->right in-order traversal(t1), and a right->root->left in-order traversal(t2), and a left->right->root postorder traversal (t3) and a right->left->root postorder traversal(t4)
the tree is symmetric iff the results of t1==t2 && t3==t4
It's not the most efficient way, but it's easy to implement