Leetcode 108. Convert Sorted Array to Binary Search Tree

Description: Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.

A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

Link: 108. Convert Sorted Array to Binary Search Tree

Examples:

Example 1:

Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:

Example 2:

Input: nums = [1,3]
Output: [3,1]
Explanation: [1,3] and [3,1] are both a height-balanced BSTs.

思路: 首先理解一下二叉搜索树的概念，它或者是一棵空树，或者是具有下列性质的二叉树： 若它的左子树不空，则左子树上所有结点的值均小于它的根节点的值； 若它的右子树不空，则右子树上所有结点的值均大于它的根结点的值； 它的左、右子树也分别是二叉搜索树。所以我们发现二叉搜索树也是被递归定义的，左右子树都是二叉搜索树。balanced强调左右子树的节点均衡，所以根节点就是sorted list 的中间节点，左右子树对应的sorted list就是以根节点切分的两部分，以此递归即可，类似之前先中序，后中序构建二叉树。

class Solution(object):
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if not nums: return None
        i = int(len(nums)/2)
        root = TreeNode(nums[i])
        root.left = self.sortedArrayToBST(nums[:i])
        root.right = self.sortedArrayToBST(nums[i+1:])
        return root

类似的题目：109. Convert Sorted List to Binary Search Tree

将sorted array 换成了sorted linkedlist, 所以第一步将链表化为array，然后一样的code.

class Solution(object):
    def sortedListToBST(self, head):
        """
        :type head: ListNode
        :rtype: TreeNode
        """
        if head is None: return None
        nums = []
        node = head
        while node:
            nums.append(node.val)
            node = node.next
        return self.sortedArrayToBST(nums)
    
    def sortedArrayToBST(self, nums):
        if not nums: return None
        i = int(len(nums)/2)
        root = TreeNode(nums[i])
        root.left = self.sortedArrayToBST(nums[:i])
        root.right = self.sortedArrayToBST(nums[i+1:])
        return root  

日期: 2021-03-15