为什么该程序的空间复杂度为O（h）？ 其中h是btree的高度

Question

So this program determines the symmetry of a btree. What is confusing me is that checkSymmetric is called twice in the same line. So wouldnt that mean we have to add two new stack frames to our callstack for each call to checkSymmetric? If thats the case shouldnt we have O(2^h) space complexity?

public static boolean isSymmetric(BinaryTreeNode<Integer> tree) {
return tree == null || checkSymmetric(tree.left, tree.right);
}

private static boolean checkSymmetric(BinaryTreeNode<Integer> subtree0,
                                    BinaryTreeNode<Integer> subtree1) {
if (subtree0 == null && subtree1 == null) {
  return true;
} else if (subtree0 != null && subtree1 != null) {
  return subtree0.data == subtree1.data
      && checkSymmetric(subtree0.left, subtree1.right)
      && checkSymmetric(subtree0.right, subtree1.left);
}
// One subtree is empty, and the other is not.
return false;
}

Answer 1

Note that we'll help, but we won't actually do your homework for you.

What is confusing me is that checkSymmetric is called twice in the same line. So wouldnt that mean we have to add two new stack frames to our callstack for each call to checkSymmetric?

No, because the two calls are sequential, not parallel. All resources scoped to the execution of one call are, by definition, released before the second call -- they are not all held at the same time. That certainly includes all the stack frames involved.

If thats the case shouldnt we have O(2^h) space complexity?

It is not the case, so what does that tell you about the space complexity?