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【leetcode】108. Convert Sorted Array to Binary Search Tree

tree return search col element left node etc lock

題目如下:

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     /    -3   9
   /   /
 -10  5

解題思路:我的方法是取nums的中間元素作為根節點,nums的左半部分為左子樹,右半部分為右子樹,同時繼續對左右子樹分別遞歸。

代碼如下:

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

class Solution(object):
    def recursive(self,node,nums):
        mid 
= len(nums)/2 left_num = nums[:mid] if len(left_num) > 0: node.left = TreeNode(left_num[len(left_num)/2]) self.recursive(node.left,left_num) right_num = nums[mid+1:] if len(right_num) > 0: node.right = TreeNode(right_num[len(right_num)/2]) self.recursive(node.right,right_num)
def sortedArrayToBST(self, nums): """ :type nums: List[int] :rtype: TreeNode """ if len(nums) == 0: return None root = TreeNode(nums[len(nums)/2]) self.recursive(root,nums) return root

【leetcode】108. Convert Sorted Array to Binary Search Tree