109. 有序链表转换二叉搜索树

原题

https://leetcode.cn/problems/convert-sorted-list-to-binary-search-tree/description/

思路

遍历 + 递归

复杂度

时间：O(n)
空间：O(n)

Python代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# 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 sortedListToBST(self, head: Optional[ListNode]) -> Optional[TreeNode]:
        l = []
        while head:
            l.append(head.val)
            head = head.next 

        # 将列表转为二叉搜索树
        def helper(l):
            if len(l) == 0:
                return None 
            idx = len(l) // 2 
            root = TreeNode(l[idx])
            root.left = helper(l[:idx])
            root.right = helper(l[idx+1:])
            return root 
        
        return helper(l)
   

Go代码

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sortedListToBST(head *ListNode) *TreeNode {
	l := []int{}
	for head != nil {
		l = append(l, head.Val)
		head = head.Next
	}
	// 将切片转为二叉搜索树
	var helper func([]int) *TreeNode
	helper = func(l []int) *TreeNode {
		if len(l) == 0 {
			return nil
		}
		idx := len(l) / 2
		root := &TreeNode{Val: l[idx]}
		root.Left = helper(l[:idx])
		root.Right = helper(l[idx+1:])
		return root
	}

	return helper(l)
}