Approach: Slow / fast pointers

def middleNode(self, head):
    slow = fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    return slow

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution:
    def middleNode(self, head):
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow

import java.util.*;
class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
    }
}
class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}

#include <bits/stdc++.h> using namespace std;
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x): val(x), next(nullptr) {
    }
};
class Solution {
    public: ListNode* middleNode(ListNode* head) {
        ListNode* slow = head;
        ListNode* fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }
};

/** Node shape: { val, next }. Build/return plain objects or use this ctor. */ function ListNode(val, next) {
    this.val = (val === undefined ? 0 : val);
    this.next = (next === undefined ? null : next);
}
/**
 * @param {ListNode} head
 * @return {ListNode}
 */ var middleNode = function(head) {
    let slow = head, fast = head;
    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
    }
    return slow;
};

Middle of the Linked List

Example 1

Example 2

Constraints