[bhyun-kim.py, 김병현] Week 7 Solutions

binary-tree-level-order-traversal/bhyun-kim.py

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+"""
+102. Binary Tree Level Order Traversal
+https://leetcode.com/problems/binary-tree-level-order-traversal/
+"""
+
+from typing import List, Optional
+
+
+# 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
+
+
+"""
+Solution 
+    Breadth First Search (BFS) using Queue
+
+    The problem is asking to return the node values at each level of the binary tree.
+    To solve this problem, we can use BFS algorithm with a queue.
+    We will traverse the tree level by level and store the node values at each level.
+
+    1. Initialize an empty list to store the output.
+    2. Initialize an empty queue.
+    3. Add the root node to the queue.
+    4. While the queue is not empty, do the following:
+        - Get the size of the queue to know the number of nodes at the current level.
+        - Initialize an empty list to store the node values at the current level.
+        - Traverse the nodes at the current level and add the node values to the list.
+        - If the node has left or right child, add them to the queue.
+        - Decrease the level size by 1.
+        - Add the list of node values at the current level to the output.
+    5. Return the output.
+
+Time complexity: O(N)
+    - We visit each node once
+
+Space complexity: O(N)
+    - The maximum number of nodes in the queue is the number of nodes at the last level
+    - The maximum number of nodes at the last level is N/2
+    - The output list stores the node values at each level which is N
+    - Thus, the space complexity is O(N)
+
+"""
+
+
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        if root is None:
+            return
+        output = []
+        queue = []
+        queue.append(root)
+
+        while len(queue) > 0:
+            level_size = len(queue)
+            level_output = []
+
+            while level_size > 0:
+                node = queue.pop(0)
+                level_output.append(node.val)
+
+                if node.left is not None:
+                    queue.append(node.left)
+                if node.right is not None:
+                    queue.append(node.right)
+
+                level_size -= 1
+
+            output.append(level_output)
+
+        return output

lowest-common-ancestor-of-a-binary-search-tree/bhyun-kim.py

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+"""
+235. Lowest Common Ancestor of a Binary Search Tree
+https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
+"""
+
+
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+
+"""
+Solution:
+    - If both p and q are greater than the root, the lowest common ancestor is in the right subtree.
+    - If both p and q are less than the root, the lowest common ancestor is in the left subtree.
+    - Otherwise, the root is the lowest common ancestor.
+
+Time complexity: O(N)
+    - The function is called recursively for each node
+Space complexity: O(N)
+    - Maximum depth of the recursion is the height of the tree
+"""
+
+
+class Solution:
+    def lowestCommonAncestor(
+        self, root: "TreeNode", p: "TreeNode", q: "TreeNode"
+    ) -> "TreeNode":
+        if p.val > root.val and q.val > root.val:
+            return self.lowestCommonAncestor(root.right, p, q)
+
+        elif p.val < root.val and q.val < root.val:
+            return self.lowestCommonAncestor(root.left, p, q)
+        else:
+            return root

remove-nth-node-from-end-of-list/bhyun-kim.py

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+"""
+19. Remove Nth Node From End of List
+https://leetcode.com/problems/remove-nth-node-from-end-of-list/
+"""
+
+
+from typing import Optional
+
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+"""
+Solution 1: 
+Two Pass Algorithm 1 
+    - First pass: Count the number of nodes and store the values in the list
+    - Second pass: Build the new list without the Nth node from the end
+
+Time complexity: O(N)
+    - Two passes are required
+Space complexity: O(N)
+    - The list stores the values of the nodes
+    - The new nodes are created to build the new list
+"""
+
+
+class Solution1:
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        vals = []
+
+        while head:
+            vals.append(head.val)
+            head = head.next
+
+        dummy_node = ListNode()
+        tail = dummy_node
+        idx_to_remove = len(vals) - n
+        vals = vals[:idx_to_remove] + vals[idx_to_remove + 1 :]
+
+        for v in vals:
+            tail.next = ListNode(val=v)
+            tail = tail.next
+
+        return dummy_node.next
+
+
+"""
+Solution 2:
+Reference: 
+    [1] https://leetcode.com/problems/remove-nth-node-from-end-of-list/editorial/
+    [2] https://www.algodale.com/problems/remove-nth-node-from-end-of-list/
+One Pass Algorithm
+    - Use two pointers to maintain a gap of n nodes between them
+    - When the first pointer reaches the end, the second pointer will be at the Nth node from the end
+
+Time complexity: O(N)
+    - Only one pass is required
+Space complexity: O(1)
+    - No extra space is required
+"""
+
+
+class Solution2:
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        dummy = ListNode()
+        dummy.next = head
+        first = dummy
+        second = dummy
+
+        for _ in range(n + 1):
+            first = first.next
+
+        while first:
+            first = first.next
+            second = second.next
+
+        second.next = second.next.next
+
+        return dummy.next

reorder-list/bhyun-kim.py

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+"""
+143. Reorder List
+https://leetcode.com/problems/reorder-list/
+"""
+
+from typing import Optional
+
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+"""
+Solution:
+    To reorder the linked list, we can follow these steps:
+    First, find the middle of the linked list using the slow and fast pointers.
+    Reverse the second half of the linked list.
+    Merge the first half and the reversed second half.
+
+    1. Find the middle of the linked list using the slow and fast pointers
+        - Initialize the slow and fast pointers to the head of the linked list
+        - Move the slow pointer by one step and the fast pointer by two steps 
+          until the fast pointer reaches the end of the linked list.
+    2. Reverse the second half of the linked list
+        - Initialize the prev and curr pointers to None and the middle node, respectively
+        - Iterate through the second half of the linked list
+            - Store the next node of the current node
+            - Reverse the current node
+            - Move the prev and curr pointers to the next nodes
+    3. Merge the first half and the reversed second half
+        - Initialize the first and second pointers to the head and the reversed second half, respectively
+        - Iterate through the linked list
+            - Store the next nodes of the first and second pointers
+            - Update the next pointers of the first and second pointers
+            - Move the first and second pointers to the next nodes
+
+Time complexity: O(N)
+    - We iterate through the linked list to find the middle node and reverse the second half
+    - The time complexity is O(N)
+
+Space complexity: O(1)
+    - We only use constant extra space for the pointers
+    - The space complexity is O(1)
+
+"""
+
+
+class Solution:
+    def reorderList(self, head: Optional[ListNode]) -> None:
+        """
+        Do not return anything, modify head in-place instead.
+        """
+        if not head or not head.next:
+            return
+
+        slow, fast = head, head
+        while fast and fast.next:
+            slow = slow.next
+            fast = fast.next.next
+
+        prev, curr = None, slow
+        while curr:
+            next_temp = curr.next
+            curr.next = prev
+            prev = curr
+            curr = next_temp
+
+        first, second = head, prev
+        while second.next:
+            tmp1, tmp2 = first.next, second.next
+            first.next = second
+            second.next = tmp1
+            first, second = tmp1, tmp2

validate-binary-search-tree/bhyun-kim.py

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+"""
+98. Validate Binary Search Tree
+https://leetcode.com/problems/validate-binary-search-tree/
+
+
+"""
+from typing import Optional
+
+
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+
+"""
+Solution1:
+    Recursion
+    In the given problem, the subtree of a node has a range of values according to the previous nodes. 
+    Thus, we can define a function that checks the validity of the subtree of a node with the range of values.
+
+    - Define a function that checks the validity of the subtree of a node with the range of values
+    - Check the validity of the left subtree and the right subtree
+        - with the range of values that the left subtree and the right subtree should have
+        - If left < root < right, the subtree is valid
+    - If the left subtree and the right subtree are valid, call the function recursively for the left and right subtrees.
+        - before calling the function, update the range of values for the left and right subtrees
+    - If the left subtree and the right subtree are valid, return True
+
+Time complexity: O(N)
+    - The function is called recursively for each node
+
+Space complexity: O(N)
+    - The function stores the range of values for each node
+"""
+
+
+class Solution1:
+    def isValidBST(self, root: Optional[TreeNode]) -> bool:
+        maximum = float("inf")
+        minimum = float("-inf")
+        return self.isValidSubTree(root, maximum, minimum)
+
+    def isValidSubTree(self, root, maximum, minimum):
+
+        if root is None:
+            return True
+
+        if not minimum < root.val < maximum: 
+            return False 
+
+        return self.isValidSubTree(
+            root.left, root.val, minimum
+            ) and self.isValidSubTree(
+            root.right, maximum, root.val 
+            )