[LeetCode] 108. Convert Sorted Array to Binary Search Tree

Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.

A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

Example 1:

Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:

Example 2:

Input: nums = [1,3]
Output: [3,1]
Explanation: [1,3] and [3,1] are both a height-balanced BSTs.

Constraints:

• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• nums is sorted in a strictly increasing order.

将有序数组转化为二叉搜索树。

给你一个整数数组 nums ，其中元素已经按 升序 排列，请你将其转换为一棵 高度平衡 二叉搜索树。

高度平衡 二叉树是一棵满足「每个节点的左右两个子树的高度差的绝对值不超过 1 」的二叉树。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree
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题目即是题意。只要输出一个有效的BST即可。此题可以跟109题一起做，要求很接近但是做法不太一样。

这个题可以用递归做。因为是BST的关系，而且input是有序数组，如果熟练，就会想到中序遍历BST的output就是一个有序数组。思路是找到数组的mid元素，即是找到了BST的根节点。小于根节点的节点是左孩子，反之是右孩子。但是注意中序遍历的结果并不能确定唯一的BST。代码如下

时间O(n)

空间O(n)

Java实现

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        // corner case
        if (nums == null || nums.length == 0) {
            return null;
        }
        return helper(nums, 0, nums.length - 1);
    }

    private TreeNode helper(int[] nums, int left, int right) {
        if (left > right)
            return null;
        int mid = left + (right - left) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = helper(nums, left, mid - 1);
        root.right = helper(nums, mid + 1, right);
        return root;
    }
}

 

JavaScript实现

/**
 * @param {number[]} nums
 * @return {TreeNode}
 */
var sortedArrayToBST = function (nums) {
    if (nums == null || nums.length == 0) return null;
    return helper(nums, 0, nums.length - 1);
}

var helper = function (nums, low, high) {
    if (low > high) return null;
    var mid = parseInt((high + low) / 2);
    var root = new TreeNode(nums[mid]);
    root.left = helper(nums, low, mid - 1);
    root.right = helper(nums, mid + 1, high);
    return root;
}

 

相关题目

108. Convert Sorted Array to Binary Search Tree

109. Convert Sorted List to Binary Search Tree

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LeetCode 题目总结