108. Convert Sorted Array to Binary Search Tree

最新推荐文章于 2022-02-19 22:56:38 发布

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最新推荐文章于 2022-02-19 22:56:38 发布

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本文介绍了一种将已排序的数组转换为高度平衡二叉搜索树的方法。平衡二叉搜索树是一种特殊的二叉树，其每个节点的左右子树的深度相差不超过1。文章提供了一个具体的例子，通过递归地选取中间元素作为根节点来构建平衡树。

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题目描述

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:

在这里插入图片描述

方法思路

class Solution {
    public TreeNode sortedArrayToBSTHelp(int left, int right, int[]nums){
        if(left > right)
            return null;
        int mid = (left + right) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        if(left == right)
            return node;
        node.left = sortedArrayToBSTHelp(left, mid - 1, nums);
        node.right = sortedArrayToBSTHelp(mid + 1, right, nums);
        return node;
    }
    
    public TreeNode sortedArrayToBST(int[] nums) {
        return sortedArrayToBSTHelp(0, nums.length - 1, nums);
    }
}