给定一个单链表,其中的元素按升序排序,将其转换为高度平衡的二叉搜索树。
本题中,一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。
示例:
给定的有序链表: [-10, -3, 0, 5, 9],
一个可能的答案是:[0, -3, 9, -10, null, 5], 它可以表示下面这个高度平衡二叉搜索树:
0
/ \
-3 9
/ /
-10 5
这个题和有序数组转换二叉搜索数基本思路一样,都是找中点作为根节点然后递归,有序数组找中点会简单一些,直接二分即可,有序链表找中点需要利用快慢指针,慢指针走一步,快指针走两步,快指针走到头慢指针指向的就是中点。
C++源代码:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* sortedListToBST(ListNode* head) {
return helper(head, NULL);
}
TreeNode* helper(ListNode* head, ListNode* tail){
if (head == tail) return NULL;
ListNode *slow = head, *fast = head;
while(fast!=tail && fast->next!=tail){
slow = slow->next;
fast = fast->next->next;
}
TreeNode *cur = new TreeNode(slow->val);
cur->left = helper(head, slow);
cur->right = helper(slow->next, tail);
return cur;
}
};
python3源代码:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def sortedListToBST(self, head: ListNode) -> TreeNode:
return self.helper(head, None)
def helper(self, head, tail):
if head==tail:
return None
slow = head
fast = head
while fast!=tail and fast.next!=tail:
slow = slow.next
fast = fast.next.next
cur = TreeNode(slow.val)
cur.left = self.helper(head, slow)
cur.right = self.helper(slow.next, tail)
return cur