我的有些形状的二叉树。我想将其转换为BST搜索相同形状的树。可能吗？

I have a binary tree of some shape. I want to Convert it to BST search tree of same shape. Is it possible?

我试过类似的方法 -

I tried methods like -

P.S：我一看这 - http://stackoverflow.com/questions/741900/binary-trees-question-checking-for-similar-shape.不过，这很容易比较2 BST的相似性的形状。

P.S.: I had a look at this - http://stackoverflow.com/questions/741900/binary-trees-question-checking-for-similar-shape. But It's easy to compare 2 BST's for similarity in shape.

推荐答案

简短的回答是：你不能。一个BST要求节点按照留下的规则＆LT; =电流I对。在这个例子中，你链接：http://upload.wikimedia.org/wikipedia/commons/f/f7/Binary_tree.svg,如果你尝试建立与相同十八一个BST你会发现，你不能。

The short answer is: you can't. A BST requires that the nodes follow the rule left <= current < right. In the example you linked: http://upload.wikimedia.org/wikipedia/commons/f/f7/Binary_tree.svg, if you try and build a BST with the same shap you'll find that you can't.

不过，如果你想舒展BST的定义，它允许左＆LT; =电流I =权（注意，这里电流I =右是允许的，作为并列的严格定义）就可以了。排序的所有元素，并把它们粘在数组中。现在做一个中序遍历，在节点与您的阵列中的每个元素替换值。下面是一些伪code：

However if you want to stretch the definition of a BST so that it allows left <= current <= right (notice that here current <= right is allowed, as apposed to the stricter definition) you can. Sort all the elements and stick them in an array. Now do an in-order traversal, replacing the values at nodes with each element in your array. Here's some pseudo code:

// t is your non-BST tree, a is an array containing the sorted elements of t, i is the current index into a
index i = 0
create_bst(Tree t, Array a)
{
  if(t is NIL)
    return;
  create_bst(t->left, a)
  t->data = a[i]
  i++
  create_bst(t->right, a)
}

结果不会但是真正的BST。如果你想有一个真正的BST这是尽量接近原来的形状成为可能，那么你再次把元素的有序数组，但这个时候将它们插入到BST。在您将它们插入的顺序由原树的子树的大小来定义。下面是一些伪code：

The result won't be a true BST however. If you want a true BST that's as close to the original shape as possible, then you again put the elements in a sorted array but this time insert them into a BST. The order in which you insert them is defined by the sizes of the subtrees of the original tree. Here's some pseudo-code:

// left is initially set to 0
create_true_bst(Tree t, BST bt, array a, index left)
{
  index i = left + left_subtree(t)->size
  bt->insert(a[i])
  if(left_subtree(t)->size != 0)
  {
    create_true_bst(t->left, bt, a, left)
    create_true_bst(t->right, bt, a, i + 1)
  }
}

这将不能保证形状是一样的但是。

This won't guarantee that the shape is the same however.