leetcode刷题题解——108. 将有序数组转换为二叉搜索树

public TreeNode sortedArrayToBST(int[] nums) {
    return sortedArrayToBST(nums,0,nums.length);
}

private TreeNode sortedArrayToBST(int[] nums,int low,int high){
    if (low>=high) return null;
    int middle = (low+high)>>1;
    TreeNode root = new TreeNode(nums[middle]);
    root.left = sortedArrayToBST(nums,low,middle);
    root.right = sortedArrayToBST(nums,middle+1,high);
    return root;
}

思路：既然要平衡，那么显然，中间节点作为根节点就很平衡了
然后进一步划分左右子树的区间，递归左右子树，即可完成该题目