题解 | #将升序数组转化为平衡二叉搜索树# | Rust

/**
 * #[derive(PartialEq, Eq, Debug, Clone)]
 * pub struct TreeNode {
 *     pub val: i32,
 *     pub left: Option<Box<TreeNode>>,
 *     pub right: Option<Box<TreeNode>>,
 * }
 *
 * impl TreeNode {
 *     #[inline]
 *     fn new(val: i32) -> Self {
 *         TreeNode {
 *             val: val,
 *             left: None,
 *             right: None,
 *         }
 *     }
 * }
 */
struct Solution{

}

impl Solution {
    fn new() -> Self {
        Solution{}
    }

    /**
    * 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
    */
    pub fn sortedArrayToBST(&self, nums: Vec<i32>) -> Option<Box<TreeNode>> {
        return Solution::dfs(self,&nums, 0, nums.len() as i32 -1);
    }
    fn dfs(&self, nums:& Vec<i32>, b:i32, e:i32) ->Option<Box<TreeNode>> {
        if b > e {
            return None;
        }
        let mid = b + (e-b)/2 + (e-b)%2;
        let mut node = Box::new(TreeNode::new(nums[mid as usize]));
        node.left = Solution::dfs(self, nums, b, mid-1);
        node.right = Solution::dfs(self, nums, mid+1, e);
        return Some(node);
    }
}