leetcode——108. 将有序数组转换为二叉搜索树

好开心，我终于独立完成了！！！！

class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        if len(nums)==0:
            return None
        i=len(nums)//2
        root=TreeNode(nums[i])
        root.left=self.sortedArrayToBST(nums[:i])
        root.right=self.sortedArrayToBST(nums[i+1:])
        return root

执行用时 :88 ms, 在所有 python3 提交中击败了77.09%的用户

内存消耗 :16.3 MB, 在所有 python3 提交中击败了5.59%的用户

 

——2019.11.15

 

---

再次复习：

public TreeNode sortedArrayToBST(int[] nums) {
        int len = nums.length;
        if(len == 0){
            return null;
        }
        int i = len/2;
        TreeNode root = new TreeNode(nums[i]);
        root.left = sortedArrayToBST(nums,0,i-1);
        root.right = sortedArrayToBST(nums,i+1,len-1);
        return root;
    }

    private TreeNode sortedArrayToBST(int[] nums, int start, int end) {
        int n = end - start + 1;
        if(n == 0){
            return null;
        }
        int i = n/2;
        TreeNode node = new TreeNode(nums[start +i]);
        node.left = sortedArrayToBST(nums,start,start +i-1);
        node.right = sortedArrayToBST(nums,start + i + 1,end);
        return node;
    }

 

 

——2020.7.2