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

    题目如下:

    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

    解题思路:我的方法是取nums的中间元素作为根节点,nums的左半部分为左子树,右半部分为右子树,同时继续对左右子树分别递归。

    代码如下:

    # 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
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  • 原文地址:https://www.cnblogs.com/seyjs/p/10625297.html
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