Second Minimum Node In a Binary Tree

Let’s break down the solution to find the second minimum value in the given special binary tree.

Understanding the Problem

In the given tree, each node with two children will have a value equal to the minimum value of its two children. We are required to find the second smallest value in this tree.

Approach

  1. Initialize Variables: Create a variable first_min to store the smallest value in the tree and a variable second_min to store the second smallest value.

  2. Traverse the Tree: Use a Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the tree. During the traversal, if we find a node’s value greater than first_min and smaller than second_min, we update second_min.

  3. Check for Second Minimum: After the traversal, if the second_min variable remains unchanged, it means there is no second minimum value in the tree, so we return -1. Otherwise, we return second_min.

Code Implementation

Here’s the code:

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class Solution:
    def findSecondMinimumValue(self, root: Optional[TreeNode]) -> int:
        # Initialize the first and second minimum values
        first_min = root.val
        second_min = float('inf')

        # Function to perform DFS
        def dfs(node):
            nonlocal second_min
            if node:
                # If the current value is greater than first_min and less than second_min, update second_min
                if first_min < node.val < second_min:
                    second_min = node.val
                # Continue DFS for left and right children
                dfs(node.left)
                dfs(node.right)

        # Start DFS from the root
        dfs(root)

        # Check if second_min was updated, else return -1
        return second_min if second_min < float('inf') else -1

Key Takeaways

  • Due to the special property of this tree, we know that the value of the root is the minimum value in the tree.
  • By performing a DFS, we can easily find the second minimum value if it exists.
  • The return value will be the second minimum value found or -1 if there’s no such value.

This code snippet efficiently solves the problem within the given constraints.

Identifying Problem Isomorphism

“Second Minimum Node In a Binary Tree” can be mapped to “Find Mode in Binary Search Tree”.

“Second Minimum Node In a Binary Tree” involves finding the second minimum value in a binary tree where each node in the tree has either 0 or 2 children. This requires a traversal of the tree to locate and compare node values.

“Find Mode in Binary Search Tree” involves finding the mode (most frequently occurring element) in a binary search tree. This also requires a traversal of the tree to count occurrences of each value.

Both problems are about traversing the binary tree and making comparisons or calculations with the node values. However, they differ in the specifics of what they’re looking for - the “Second Minimum Node In a Binary Tree” problem is looking for a second smallest value, while the “Find Mode in Binary Search Tree” problem is looking for the most common value. In terms of complexity, “Find Mode in Binary Search Tree” is slightly more complicated since it requires keeping track of frequencies of all elements.

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class Solution:
    def findSecondMinimumValue(self, root: TreeNode) -> int:
        nums = []
        nodes = [root]
        while nodes:
            t = nodes.pop()
            nums.append(t.val)
            if t.left:
                nodes.append(t.left)
            if t.right:
                nodes.append(t.right)

        if len(set(nums)) == 1:
            return -1

        return sorted(set(nums))[1]

Problem Classification

The problem falls under the Computer Science domain, more specifically, in the field of Data Structures, with a focus on Binary Trees.

‘What’ Components:

  1. A special non-empty binary tree where each node has either two or zero children.
  2. If a node has two children, the node’s value is the minimum of its two children’s values.
  3. The task is to find the second smallest value among all nodes in the tree.
  4. If no second minimum value exists, the function should return -1.

This is a search and traversal problem on binary trees where we are required to find the second smallest value in a tree with a specific structural property. It requires understanding the properties of binary trees, traversing the tree, comparing node values, and selecting the appropriate value based on the criteria given in the problem. It is a problem involving understanding and applying tree traversal algorithms and handling edge cases (no second minimum value).

Language Agnostic Coding Drills

  1. Dissection of the Code:

  • Initialization of Data Structures: The code begins by initializing an empty list, nums, and a list nodes containing the root of the tree. This is a straightforward concept used to store the values of the nodes and the nodes themselves that we are going to traverse.

  • Tree Traversal: The next concept is the tree traversal using a while loop. This loop continues as long as there are elements in the nodes list. This is a common strategy for breadth-first search (BFS) or depth-first search (DFS) on trees or graphs. In this case, it is an iterative version of DFS since nodes are popped from the end of the list.

  • Condition Checking and List Appending: The code checks whether each node has left or right children, and if so, it appends these to the nodes list. This is necessary for traversal in binary trees.

  • Set Operations: The code uses a set to eliminate duplicate values from the nums list and checks its length. This is a technique used for eliminating duplicates and checking unique elements in a list.

  • Sorting and Indexing: The code then sorts the unique values in the nums list and returns the second element. This is a method to find the second smallest value in the list.

  1. List of Identified Coding Concepts (from simple to complex):

  • List Initialization: The first step of the code, initializing the lists, is a basic concept and forms the foundation for more complex operations.

  • List Appending: Adding elements to a list is a crucial concept in Python and many other languages.

  • Tree Traversal (DFS): Traversing a tree using DFS can be a complex concept, especially for beginners. This concept involves understanding the structure of the binary tree and the nature of depth-first search.

  • Condition Checking: Condition checking using if statements is an intermediate concept. It is fundamental in controlling the flow of the program.

  • Set Operations: Converting a list to a set to eliminate duplicates and check unique values is an intermediate concept. It involves understanding different data structures and their properties.

  • Sorting and Indexing: Sorting a list and retrieving an element at a specific index is a slightly more advanced concept as it involves manipulating data structures based on certain conditions.

  1. Problem-solving Approach and Strategy:

This code solves the problem in a step-by-step manner:

  • The code first initializes two lists, nums and nodes, to store the values of the nodes and the nodes themselves, respectively.

  • The program then starts to traverse the tree, popping a node from the nodes list, and appending its value to the nums list.

  • For each node popped, if the node has left or right children, these children are appended to the nodes list, continuing the traversal.

  • After the tree has been completely traversed, the nums list is converted into a set to remove duplicates and check the count of unique values.

  • If there is only one unique value, the function returns -1 as per the problem statement.

  • Otherwise, the unique values are sorted, and the second value (i.e., the second minimum) is returned.

Each of these steps or drills contributes to the overall solution by breaking down the problem into manageable tasks that perform specific functions, and when integrated, they solve the problem in an effective and efficient manner.

Targeted Drills in Python

  1. Python Coding Drills for Identified Concepts:
  • List Initialization: A basic Python list can be initialized as follows:
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nums = []
nodes = [root]
  • List Appending: Adding elements to a Python list can be done using the append function:
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nums.append(value)
nodes.append(node)
  • Tree Traversal (DFS): Traversing a tree using DFS involves visiting the root, traversing the left subtree, and then the right subtree. Here’s an iterative implementation:
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nodes = [root]
while nodes:
    node = nodes.pop()
    # Process node
    if node.right:
        nodes.append(node.right)
    if node.left:
        nodes.append(node.left)
  • Condition Checking: In Python, condition checking is done using if statements:
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if node.left:
    # Do something
  • Set Operations: Converting a list to a set in Python, and checking the length of the set:
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unique_nums = set(nums)
count = len(unique_nums)
  • Sorting and Indexing: Sorting a list in Python and retrieving an element at a specific index:
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sorted_nums = sorted(unique_nums)
second_min = sorted_nums[1]
  1. Problem-specific Concepts:

  • The tree traversal in this problem is a form of Depth-First Search, but with no particular order between left and right subtrees because we want to visit all nodes.

  • The checking and handling of the case where all nodes have the same value is specific to this problem. If the length of the set is 1, it means all nodes have the same value and there’s no second minimum value, hence we return -1.

  1. Integration of the Drills:

These drills can be combined to solve the problem in the following manner:

  • Begin by initializing the nums and nodes lists, with nodes initially containing the root of the tree.

  • Start a while loop that continues as long as nodes has elements. In each iteration, pop a node from nodes, append its value to nums, and if it has left or right children, append them to nodes.

  • After the tree is completely traversed, convert nums to a set to remove duplicates and check its length. If the length is 1, return -1.

  • If the length of the set is more than 1, sort it, and return the second element.

Each of these drills, when assembled in the correct order, gives the solution to the problem.