【二叉树】108.将有序数组转换为二叉搜索树

题目

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
        return self.build(nums, 0, len(nums)-1)
    
    def build(self, nums: List[int], start_idx: int, end_idx: int):
        if start_idx > end_idx:
            return None
        mid = start_idx + (end_idx - start_idx) // 2
        root = TreeNode(nums[mid])
        left = self.build(nums, start_idx, mid-1)
        right = self.build(nums, mid+1, end_idx)
        root.left = left
        root.right = right
        return root

Java

法1:递归

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums.length == 0) {
            return null;
        } else if (nums.length  == 1) {
            return new TreeNode(nums[0]);
        }

        return build(nums, 0, nums.length - 1);
    }

    public TreeNode build(int[] nums, int start, int end) {
        if (start > end) {
            return null;
        } else if (start == end) {
            return new TreeNode(nums[start]);
        }
        int mid = start + (end - start) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = build(nums, start, mid - 1);
        root.right = build(nums, mid + 1, end);
        return root;
    }
}