题解 | #将升序数组转化为平衡二叉搜索树#
将升序数组转化为平衡二叉搜索树
https://www.nowcoder.com/practice/7e5b00f94b254da599a9472fe5ab283d
/*
* function TreeNode(x) {
* this.val = x;
* this.left = null;
* this.right = null;
* }
*/
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param nums int整型一维数组
* @return TreeNode类
*/
function sortedArrayToBST( nums ) {
if(nums.length === 0) {
return null
}
function tree(nums, left, right) {
if(left > right) return null
// const mid = Math.floor((right - left) / 2); // 向低位取, 也是正确的, 只是案例给的是向高位取的结果
const mid = Math.ceil((left + right) / 2); // 向高位取
const root = new TreeNode(nums[mid])
root.left = tree(nums, left, mid - 1)
root.right = tree(nums, mid + 1, right)
return root;
}
return tree(nums, 0, nums.length - 1)
// write code here
}
module.exports = {
sortedArrayToBST : sortedArrayToBST
};


