109. 有序链表转换二叉搜索树

• 109. 有序链表转换二叉搜索树

  • 题目描述

  • 方法 1：递归

    • 思路

    • 复杂度分析

    • 代码(JavaScript/C++/Python)

  • 方法 2：空间换时间

    • 思路

    • 复杂度分析

    • 代码(JavaScript/C++)

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：

      0
     / \
   -3   9
   /   /
 -10  5

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/convert-sorted-list-to-binary-search-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

方法 1：递归

思路

• 先用快慢指针找到中间节点

• 分治构建平衡二叉树

复杂度分析

• 时间复杂度：$O(NlogN)$，N 为链表长度。

• 空间复杂度：$O(logN)$，N 为链表长度。

代码(JavaScript/C++/Python)

JavaScript Code

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {ListNode} head
 * @return {TreeNode}
 */
var sortedListToBST = function (head, tail = null) {
  if (!head || head === tail) return null;

  let slow = head,
    fast = head;
  while (fast !== tail && fast.next !== tail) {
    slow = slow.next;
    fast = fast.next.next;
  }

  const root = new TreeNode(slow.val);
  root.left = sortedListToBST(head, slow);
  root.right = sortedListToBST(slow.next, tail);
  return root;
};

C++ Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        if (head == nullptr) return nullptr;
        return sortedListToBST(head, nullptr);
    }
    TreeNode* sortedListToBST(ListNode* head, ListNode* tail) {
        if (head == tail) return nullptr;

        ListNode* slow = head;
        ListNode* fast = head;

        while (fast != tail && fast->next != tail) {
            slow = slow->next;
            fast = fast->next->next;
        }

        TreeNode* root = new TreeNode(slow->val);
        root->left = sortedListToBST(head, slow);
        root->right = sortedListToBST(slow->next, tail);
        return root;
    }
};

Python Code

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sortedListToBST(self, head):
        """
        :type head: ListNode
        :rtype: TreeNode
        """
        if not head: return None
        prev, slow, fast = None, head, head

        while fast and fast.next:
            prev = slow
            slow = slow.next
            fast = fast.next.next

        root = TreeNode(slow.val)
        if slow == fast: return root

        if prev: prev.next = None
        root.left = self.sortedListToBST(head)
        root.right = self.sortedListToBST(slow.next)
        return root

方法 2：空间换时间

思路

由于寻找链表中点的时间复杂度是 $O(N)$，如果事先使用数组将链表的值存储起来，寻找中点就变成了 $O(1)$ 时间的操作，代价则是 $O(N)$ 的额外空间，问题转换成了将有序数组转换成搜索二叉树。

复杂度分析

• 时间复杂度：$O(N)$，N 为链表长度。

• 空间复杂度：$O(N)$，N 为链表长度。

代码(JavaScript/C++)

JavaScript Code

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {ListNode} head
 * @return {TreeNode}
 */
var sortedListToBST = function (head) {
  if (!head) return null;

  const nodes = [];
  while (head) {
    nodes.push(head.val);
    head = head.next;
  }

  return sortedArrayToBST(nodes, 0, nodes.length);

  // ********************************************
  function sortedArrayToBST(array, start, end) {
    if (start >= end) return null;

    const mid = (((end - start) >> 1) >> 0) + start;
    const root = new TreeNode(array[mid]);
    root.left = sortedArrayToBST(array, start, mid);
    root.right = sortedArrayToBST(array, mid + 1, end);
    return root;
  }
};

C++ Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        vector<int> nodes;
        while (head != nullptr) {
            nodes.push_back(head->val);
            head = head->next;
        }
        return sortedListToBST(nodes, 0, nodes.size());
    }
    TreeNode* sortedListToBST(vector<int>& nodes, int start, int end) {
        if (start >= end) return nullptr;

        int mid = (end - start) / 2 + start;
        TreeNode* root = new TreeNode(nodes[mid]);
        root->left = sortedListToBST(nodes, start, mid);
        root->right = sortedListToBST(nodes, mid + 1, end);
        return root;
    }
};

更多题解可以访问：https://github.com/suukii/91-days-algorithm