是否可以从未排序的数组中高效地创建平衡的二进制搜索树而无需对数组进行排序？

Question

The title says it all.

I see that I can create a binary search tree out of an unsorted array rather easily.

If root is null, set 1st array value to the root
current = root
for each value in the array:
  while current not null
     If arrays value >= current value
          if root.left is null, set array value to current.right
          else current = current.right and continue
     Else if arrays value < current value
          if current.left is null, set array value to current.left
          else current = current.left
return root;

And can also create a balanced binary search tree out of an ordered array easily.

Get the Middle of the array and make it root.
Recursively do same for left half and right half.
      Get the middle of left half and make it left child of the root created in step 1.
      Get the middle of right half and make it right child of the root created in step 1.

But is there an efficient way to create a balanced binary search tree from an unsorted array just as easily without changing the array / copying the array, ect.

Answer 1

Your second approach is likely to be the simplest of all if you have no libraries at hand. It's also very efficient if you use a good sorting algorithm (asymptotically optimal with a very low constant factor).

Your first approach is not really efficient, because the tree can become unbalanced. You can however insert all elements into a self-balancing binary search tree one by one, in any order. This also needs time O(n log n) , like the second approach.

Of course you won't be able to do it faster than that, because then you would have essentially sorted the array in o(n log n) using only comparisons, which is impossible .