LC Problems

116. Populating Next Right Pointers in Each Node/116. Populating Next Right Pointers in Each Node.py

@@ -28,16 +28,13 @@ def __init__(self, val, left, right, next):
         self.next = next
 """
 class Solution:
-    def connect(self, root: 'Node') -> 'Node':
-        if not root:
-            return None
-        if root.left:
-            root.left.next = root.right
+    def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
+        if not root: return None
         if root.right:
+            if root.left:
+                root.left.next = root.right
             if root.next:
                 root.right.next = root.next.left
-            else:
-                root.right.next = None
         self.connect(root.left)
         self.connect(root.right)
         return root

1325. Delete Leaves With a Given Value/1325. Delete Leaves With a Given Value.py

@@ -0,0 +1,48 @@
+"""
+Given a binary tree root and an integer target, delete all the leaf nodes with value target.
+
+Note that once you delete a leaf node with value target, if its parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you cannot).
+
+
+
+Example 1:
+
+
+
+Input: root = [1,2,3,2,null,2,4], target = 2
+Output: [1,null,3,null,4]
+Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left).
+After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).
+Example 2:
+
+
+
+Input: root = [1,3,3,3,2], target = 3
+Output: [1,3,null,null,2]
+Example 3:
+
+
+
+Input: root = [1,2,null,2,null,2], target = 2
+Output: [1]
+Explanation: Leaf nodes in green with value (target = 2) are removed at each step.
+
+
+Constraints:
+
+The number of nodes in the tree is in the range [1, 3000].
+1 <= Node.val, target <= 1000
+"""
+
+
+# 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 removeLeafNodes(self, root: Optional[TreeNode], target: int) -> Optional[TreeNode]:
+        if root.left: root.left = self.removeLeafNodes(root.left, target)
+        if root.right: root.right = self.removeLeafNodes(root.right, target)
+        return None if not root.left and not root.right and root.val == target else root

3217. Delete Nodes From Linked List Present in Array/3217. Delete Nodes From Linked List Present in Array.py

@@ -0,0 +1,79 @@
+"""
+You are given an array of integers nums and the head of a linked list. Return the head of the modified linked list after removing all nodes from the linked list that have a value that exists in nums.
+
+
+
+Example 1:
+
+Input: nums = [1,2,3], head = [1,2,3,4,5]
+
+Output: [4,5]
+
+Explanation:
+
+
+
+Remove the nodes with values 1, 2, and 3.
+
+Example 2:
+
+Input: nums = [1], head = [1,2,1,2,1,2]
+
+Output: [2,2,2]
+
+Explanation:
+
+
+
+Remove the nodes with value 1.
+
+Example 3:
+
+Input: nums = [5], head = [1,2,3,4]
+
+Output: [1,2,3,4]
+
+Explanation:
+
+
+
+No node has value 5.
+"""
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+class Solution:
+    def modifiedList(self, nums, head):
+        hset = set(nums)
+        temp = ListNode(-100)
+        temp.next = head
+        result = temp
+        while temp.next:
+            if temp.next.val in hset:
+                temp.next = temp.next.next
+            else:
+                temp = temp.next
+        return result.next
+
+head = ListNode(1)
+head.next = ListNode(2)
+head.next.next = ListNode(3)
+head.next.next.next = ListNode(4)
+head.next.next.next.next = ListNode(5)
+
+# Sample input for nums
+nums = [1, 2, 3]
+
+# Create an instance of the Solution class
+solution = Solution()
+
+# Call the modifiedList method, passing both nums and head
+result = solution.modifiedList(nums, head)
+
+# Print the modified linked list
+while result:
+    print(result.val)
+    result = result.next

725. Split Linked List in Parts/725. Split Linked List in Parts.py

@@ -0,0 +1,63 @@
+"""
+Given the head of a singly linked list and an integer k, split the linked list into k consecutive linked list parts.
+
+The length of each part should be as equal as possible: no two parts should have a size differing by more than one. This may lead to some parts being null.
+
+The parts should be in the order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal to parts occurring later.
+
+Return an array of the k parts.
+
+
+
+Example 1:
+
+
+Input: head = [1,2,3], k = 5
+Output: [[1],[2],[3],[],[]]
+Explanation:
+The first element output[0] has output[0].val = 1, output[0].next = null.
+The last element output[4] is null, but its string representation as a ListNode is [].
+Example 2:
+
+
+Input: head = [1,2,3,4,5,6,7,8,9,10], k = 3
+Output: [[1,2,3,4],[5,6,7],[8,9,10]]
+Explanation:
+The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.
+
+
+Constraints:
+
+The number of nodes in the list is in the range [0, 1000].
+0 <= Node.val <= 1000
+1 <= k <= 50
+"""
+
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def splitListToParts(self, head, k):
+        count, temp = 0, head
+        while temp:
+            temp = temp.next
+            count += 1
+        chunk_size = count // k
+        extra_part = count % k
+        cur = head
+        result = []
+        for i in range(k):
+            part_length = chunk_size + (1 if i < extra_part else 0)
+            part_head = cur
+            for j in range(part_length - 1):
+                if cur:
+                    cur = cur.next
+            if cur:
+                next_part = cur.next
+                cur.next = None
+                cur = next_part
+            result.append(part_head)
+        return result