LeetCode109 链表转二叉搜索树

最新推荐文章于 2025-05-16 01:45:00 发布

tigerhuli

最新推荐文章于 2025-05-16 01:45:00 发布

阅读量219

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/m0_37518259/article/details/81120892

LeetCode 专栏收录该内容

503 篇文章

订阅专栏

本文介绍如何将一个按升序排列的单链表转换成高度平衡的二叉搜索树。平衡二叉树定义为任意节点的两个子树深度差不超过1。文中提供了两种方法：一种通过中间数组辅助，另一种直接利用链表元素顺序构造。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

题源：here；完整实现：here

思路：

两种方案，一种偷懒的方案，先转换为数组再转二叉树，另一种是利用生成树的先后顺序进行二叉树的构造。

1 偷懒版

TreeNode* sortedArrayToBST(vector<int>& nums) {
	if (nums.size() == 0) return NULL;
	int mid = nums.size() / 2;
	TreeNode *root = new TreeNode(nums[mid]);
	vector<int> leftNums(nums.begin(), nums.begin() + mid);
	vector<int> rightNums(nums.begin() + mid + 1, nums.end());
	root->left = sortedArrayToBST(leftNums);
	root->right = sortedArrayToBST(rightNums);

	return root;
}

TreeNode* sortedListToBST(ListNode* head) {
	vector<int> nums;
	ListNode *curr = head;
	while (curr){
		nums.push_back(curr->val);
		curr = curr->next;
	}

	return sortedArrayToBST(nums);
}

2 改变顺序版

TreeNode *sortedListToBST2(int left, int right, ListNode *&head){
	if (!head || left > right) return NULL;
	int mid = left + (right - left) / 2;
	TreeNode *leftNode = sortedListToBST2(left, mid - 1, head);
	TreeNode *root = new TreeNode(head->val);
	head = head->next;
	TreeNode *rightNode = sortedListToBST2(mid + 1, right, head);
	root->left = leftNode;
	root->right = rightNode;

	return root;
}

TreeNode* sortedListToBST2(ListNode* head) {
	int len = 0;
	ListNode *curr = head;
	while (curr){
		len++;
		curr = curr->next;
	}

	return sortedListToBST2(0, len, head);
}