最小化二叉搜索树的高度

Question

If I am attempting to minimize the height of a Binary Search Tree, are these the correct steps?:

1) produce a sorted array from the tree 2) reconstruct the tree by adding the sorted elements into the tree inorder

Answer 1

Adding an already sorted list to a simple non-balancing binary search tree will build the theoretical Worst case for a binary search tree. The lowest-valued node is the root, every node is added to the 'right' of the node immediately preceding it in the list, and you create a tree of maximum depth, searching in O(n) time rather than O(lg n). You'dd effectively just be building an overly complicated linked-list.

Answer 2

对元素进行排序之后，可以通过将中间元素定义为根节点来重建树，然后分别从中间元素之前和之后的元素递归地构建左右子树。

Answer 3

I think if you reconstruct the tree structure before you try to insert sorted elements via inorder , the solution you provide will be right.

1. Reconstruct the tree. just like a heap.
2. Insert sorted element via inorder

For example, if the original tree is like this:

                 (5)
           (3)         (6)
      (2)      (4)
  (1)

1. Reconstruct the tree like this:

     () () () () () () 

2. Insert sorted element via inorder : 1, 2, 3, 4, 5, 6

     (4) (2) (6) (1) (3) (5) 

Answer 4

I suppose you have access to the tree and can alter it "manually". I think your balancing problem could be solved like this (pseudocode):

depth(node)
{
    if node is null, return 0;
    l = depth(left child);
    r = depth(right child);
    diff = (r - l);
    if (diff < -1) rotate right (as often as you need);
    else if (diff > 1) rotate left (as often as you need);
    return the new maximum depth of both subtrees +1;
}

I must confess, I am not very sure about this, but the idea is that you don't need the temporary array, because traversing the tree and applying the right rotations should do.