把一个排好序的数组数组转化成一颗平衡二叉查询树 [No. 79]

最新推荐文章于 2024-08-19 17:45:04 发布

最新推荐文章于 2024-08-19 17:45:04 发布 · 63 阅读

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本文探讨了如何将有序数组转换为平衡二叉搜索树，通过递归方式选择中间元素作为根节点，确保树的高度最小化，实现O(N)时间复杂度的高效转换。同时，对比了此过程与二分查找算法的相似之处，揭示了它们共同遵循的分治策略。

Solution:
If you would have to choose an array element to be the root of a balanced BST, which element would you pick? The root of a balanced BST should be the middle element from the sorted array.

You would pick the middle element from the sorted array in each iteration. You then create a node in the tree initialized with this element. After the element is chosen, what is left? Could you identify the sub-problems within the problem?

There are two arrays left — The one on its left and the one on its right. These two arrays are the sub-problems of the original problem, since both of them are sorted. Furthermore, they are subtrees of the current node’s left and right child.

The code below creates a balanced BST from the sorted array inO(N) time (Nis the number of elements in the array). Compare how similar the code is to a binary search algorithm. Both are using the divide and conquer methodology.

BinaryTree* sortedArrayToBST(int arr[], int start, int end) {
  if (start > end) return NULL;
  // same as (start+end)/2, avoids overflow.
  int mid = start + (end - start) / 2;
  BinaryTree *node = new BinaryTree(arr[mid]);
  node->left = sortedArrayToBST(arr, start, mid-1);
  node->right = sortedArrayToBST(arr, mid+1, end);
  return node;
}
 
BinaryTree* sortedArrayToBST(int arr[], int n) {
  return sortedArrayToBST(arr, 0, n-1);
}

参考： http://www.ihas1337code.com/2010/11/convert-sorted-array-into-balanced.html

这个问题常常会是其它问题的一个子问题，比如，如何把两个BST合并成一个BST，要求时间复杂度为O(N)。

大致的思路是：用in-order walk得到两组排序的数组（O(N)），合并两个数组（O(N)）,然后就是把数组转化成平衡二叉树（O(N)）。