【DFS】LeetCode 108. 将有序数组转换为二叉搜索树

题目链接

108. 将有序数组转换为二叉搜索树

思路

类似于二分搜索，定位到数组中间 mid，然后左边的子数组构成左子树，右边的子数组构成右子树，mid 处的数字构成根结点。递归构建左右子树即可，最后返回 root。

代码

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        return buildBST(nums, 0, nums.length - 1);
    }

    public TreeNode buildBST(int[] nums, int left, int right){
        if(left > right){
            return null;
        }

        int mid = (right - left) / 2 + left;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = buildBST(nums, left, mid - 1);
        root.right = buildBST(nums, mid + 1, right);

        return root;
    }
}