Leetcode: Convert Sorted Array to Binary Search Tree-优快云博客

本文介绍了一种方法，通过深度优先搜索，将一个按升序排列的数组转换为高度平衡的二叉搜索树。详细阐述了算法实现步骤，并提供了代码示例。

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Question

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

Hide Tags Tree Depth-first Search

Analysis

use binary search

Solution

v1.

The stop condition of depth-search tree is when subarray has only one or two elements.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param {integer[]} nums
    # @return {TreeNode}
    def sortedArrayToBST(self, nums):
        if len(nums)==0:
            return None

        return self.helper(nums, 0, len(nums)-1)

    def helper(self,nums,left,right):
        if right==left:
            node1 = TreeNode(nums[right])
            return node1

        if right-left==1:
            node1 = TreeNode(nums[right])
            node2 = TreeNode(nums[left])
            node2.right = node1
            return node2

        mid = (left+right)/2
        subroot  = TreeNode(nums[mid])
        subroot.left = self.helper(nums,left,mid-1)
        subroot.right = self.helper(nums,mid+1,right)
        return subroot

v2

The stop condition is changed for creating None.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param {integer[]} nums
    # @return {TreeNode}
    def sortedArrayToBST(self, nums):
        if len(nums)==0:
            return None

        return self.helper(nums, 0, len(nums)-1)

    def helper(self,nums,left,right):
        if left>right:
            return None

        mid = (left+right)/2
        subroot  = TreeNode(nums[mid])
        subroot.left = self.helper(nums,left,mid-1)
        subroot.right = self.helper(nums,mid+1,right)
        return subroot