LeetCode #109 Convert Sorted List to Binary Search Tree 有序链表转换二叉搜索树

109 Convert Sorted List to Binary Search Tree 有序链表转换二叉搜索树

Description:
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:

题目描述:
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：

思路:

递归法
使用快慢指针找到链表中点即为根节点
前一半链表为左子树, 后一半链表为右子树
时间复杂度O(nlgn), 空间复杂度O(lgn)

代码:
C++:

/**
 * 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) 
    {
        if (!head) return nullptr;
        TreeNode *root = new TreeNode(head -> val);
        if (!head -> next) return root;
        ListNode *pre = head, *p = pre -> next, *q = p -> next;
        while (q and q -> next)
        {
            pre = pre -> next;
            p = pre -> next;
            q = q -> next -> next;
        }
        pre -> next = nullptr;
        root -> val = p -> val;
        root -> left = sortedListToBST(head);
        root -> right = sortedListToBST(p -> next);
        return root;
    }
};

Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if (head == null) return null;
        if (head.next == null) return new TreeNode(head.val);
        ListNode pre = head, p = head.next, q = head.next.next;
        while (q != null && q.next != null) {
            pre = pre.next;
            p = pre.next;
            q = q.next.next;
        }
        pre.next = null;
        TreeNode root = new TreeNode(p.val);
        root.left = sortedListToBST(head);
        root.right = sortedListToBST(p.next);
        return root;
    }
}

Python:

# 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:
        def helper(head: TreeNode, tail: TreeNode) -> TreeNode:
            if head == tail:
                return
            slow, fast = head, head
            while fast != tail and fast.next != tail:
                slow, fast = slow.next, fast.next.next
            root = TreeNode(slow.val)
            root.left, root.right = helper(head,slow), helper(slow.next,tail)
            return root
        return helper(head, None)

最后编辑于：2021.04.12 17:59:20