108.将有序数组转换为二叉搜索树 双百0ms

区间分治即可

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        return dfs(nums, 0, nums.length - 1);
    }
    public TreeNode dfs(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 = dfs(nums, left, mid - 1);
        root.right = dfs(nums, mid + 1, right);
        return root;
    }
}

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来源： https://blog.csdn.net/Split_token/article/details/122144522