为什么要添加不平衡二叉搜索树O（n）？

Question

This is the implementation of add in Binary Search Tree from BST Add

private IntTreeNode add(IntTreeNode root, int value) {
        if (root == null) {
            root = new IntTreeNode(value);
        } else if (value <= root.data) {
            root.left = add(root.left, value);
        } else {
            root.right = add(root.right, value);
        }

        return root;
    }

I understand why this runs in O(log n). Here's how I analyze it. We have a tree size of n. How many cuts of 2, or half cut, will reduce this tree down to a size of 1. So we have the expression n(1/2)^x = 1 where the 1/2 represents each half cut. Solving this for x, we have log2(x) so the logn comes from search.
Here is a lecture slide from Heap that discusses runtime for an unbalanced binary search.

My question is even if the binary search tree is unbalanced, wouldn't the same strategy work for analyzing the runtime of add? How many cuts you have to make. Wouldn't the runtime still be O(log n), not O(n)? If so, can someone show the math of why it would be O(n)?

Answer 1

With an unbalanced tree:

1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          ...

Your intuition of cutting the tree in half with each operation no longer applies. This unbalanced tree is the worst case of an unbalanced binary search tree. To search for 10 at the bottom of the list, you must make 10 operations, one for each element in the tree. That is why a search operation for an unbalanced binary search tree is O (n) - this unbalanced binary search tree is equivalent to a linked list. Each operation doesn't cut off half the tree -- just the one node you've already visited.

That is why specialized versions of binary search trees, such as red-black trees and AVL trees are important: they maintain trees that are balanced well enough so that all operations - search, insert, delete -- are still O (log n).

Answer 2

The O(n) situation in a BST happens when you have either the minimum or the maximum at the top, effectively turning your BST into a linked list. Suppose you added elements as: 1, 2, 3, 4, 5 , generating your BST, which will be a linked list due to every element having only a right child . Adding 6 would have to go down right on every single node, going through all the elements, hence making the asymptotic complexity of add O(n)