convert a sorted array into a balanced binary search tree

最新推荐文章于 2021-10-25 19:38:05 发布

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#search #tree #arrays #each #扩展 #2010

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本文介绍了一种方法，通过选择排序数组的中位数元素作为根节点，将其转换为平衡二叉搜索树。这种方法利用了递归和分治策略，确保了时间复杂度为O(N)，并提供了与二分查找相似的代码实现。

Question:

Given a sorted array, and converted it into a balanced binary search tree.

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 in O(N) time (N is 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.

Node 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;
	Node node = new Node(arr[mid]);
	node.leftChild = sortedArrayToBST(arr, start, mid-1);
	node.rightChild = sortedArrayToBST(arr, mid+1, end);
	return node;
}	 
Node sortedArrayToBST(int arr[], int n) {
	return sortedArrayToBST(arr, 0, n-1);
}

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

扩展：
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)）。