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109. Convert Sorted List to Binary Search Tree

中文文档

Description

Given the head of 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 1:

Input: head = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.

Example 2:

Input: head = []
Output: []

Example 3:

Input: head = [0]
Output: [0]

Example 4:

Input: head = [1,3]
Output: [3,1]

 

Constraints:

  • The number of nodes in head is in the range [0, 2 * 104].
  • -10^5 <= Node.val <= 10^5

Solutions

Python3

# 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: ListNode) -> TreeNode:
        def buildBST(nums, start, end):
            if start > end:
                return None
            mid = (start + end) >> 1
            return TreeNode(nums[mid], buildBST(nums, start, mid - 1), buildBST(nums, mid + 1, end))
        
        nums = []
        while head:
            nums.append(head.val)
            head = head.next
        return buildBST(nums, 0, len(nums) - 1)

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        List<Integer> nums = new ArrayList<>();
        for (; head != null; head = head.next) {
            nums.add(head.val);
        }
        return buildBST(nums, 0, nums.size() - 1);
    }

    private TreeNode buildBST(List<Integer> nums, int start, int end) {
        if (start > end) {
            return null;
        }
        int mid = (start + end) >> 1;
        TreeNode root = new TreeNode(nums.get(mid));
        root.left = buildBST(nums, start, mid - 1);
        root.right = buildBST(nums, mid + 1, end);
        return root;
    }
}

C++

/**
 * 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> nums;
        for (; head != nullptr; head = head->next) {
            nums.push_back(head->val);
        }
        return buildBST(nums, 0, nums.size() - 1);
    }

private:
    TreeNode* buildBST(vector<int>& nums, int start, int end) {
        if (start > end) {
            return nullptr;
        }
        int mid = (start + end) / 2;
        TreeNode *root = new TreeNode(nums[mid]);
        root->left = buildBST(nums, start, mid - 1);
        root->right = buildBST(nums, mid + 1, end);
        return root;
    }
};

JavaScript

/**
 * 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) {
    const buildBST = (nums, start, end) => {
        if (start > end) {
            return null;
        }
        const mid = (start + end) >> 1;
        const root = new TreeNode(nums[mid]);
        root.left = buildBST(nums, start, mid - 1);
        root.right = buildBST(nums, mid + 1, end);
        return root;
    }

    const nums = new Array();
    for (; head != null; head = head.next) {
        nums.push(head.val);
    }
    return buildBST(nums, 0, nums.length - 1);
};

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