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  • 二叉搜索树 | 将有序数组转换为二叉搜索树

    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
    
    

    思路:

    其实就是构建一个二叉搜索树,对于搜索树,有三个特点:
    1、根节点左边的都比根节点小。
    2、根节点右边的都比根节点大。
    3、左孩子和右孩子都是搜索树。
    因此通过递归构建左孩子,右孩子。

    class Solution(object):
        def sortedArrayToBST(self, nums):
            """
            :type nums: List[int]
            :rtype: TreeNode
            """
            if len(nums) == 1:
                return TreeNode(nums[0])
            elif len(nums) == 0:
                return None
            else:
                mid = (len(nums)-1) / 2
                node = TreeNode(nums[mid])
                node.left = self.sortedArrayToBST(nums[:mid])
                node.right = self.sortedArrayToBST(nums[mid+1:])
                return node
    
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  • 原文地址:https://www.cnblogs.com/xmxj0707/p/10381155.html
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