Given a binary tree where each node has a value and each node's value is either equal to the value of its parent or strictly greater than the value of its parent, find the second smallest value in the tree. The smallest value will be the root node value. If no second smallest value exists, return -1.

For example, consider the following binary tree:

    2
   / \
  2   5
 / \ / \
5  7 5  7

In this case, the smallest value is 2. The next smallest value is 5. Therefore, the function should return 5.

Consider another example:

    2
   / \
  2   2
 / \ / \
2  2 2  2

Here, all node values are 2. There is no second smallest value. The function should return -1.

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• 0 <= Node.val <= 2^31 - 1
• root.val is always the smallest value in the tree.

Finding the Second Smallest Value in a Binary Tree

Problem Description

Given a binary tree where each node has a value, and each node's value is either equal to the value of its parent or strictly greater than the value of its parent, find the second smallest value in the tree. The smallest value will be the root node value. If no second smallest value exists, return -1.

For example:

    2
   / \
  2   5
 / \ / \
5  7 5  7

In this case, the smallest value is 2. The next smallest value is 5. Therefore, the function should return 5.

Consider another example:

    2
   / \
  2   2
 / \ / \
2  2 2  2

Here, all node values are 2. There is no second smallest value. The function should return -1.

Naive Approach

A straightforward approach is to traverse the entire tree, collect all unique values into a set, remove the smallest value (which is the root's value), and then return the minimum value from the remaining set, if the set is not empty. If the set is empty, it means there is no second smallest value, and we return -1.

Code Example (Python)

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def find_second_minimum_value_naive(root: TreeNode) -> int:
    unique_values = set()
    
    def traverse(node):
        if node:
            unique_values.add(node.val)
            traverse(node.left)
            traverse(node.right)

    traverse(root)
    unique_values.discard(root.val)
    
    if not unique_values:
        return -1
    else:
        return min(unique_values)

Optimal Approach

Since we know that the root node contains the smallest value, we can optimize the search. We only need to find the smallest value that is greater than the root's value. We can perform a depth-first search (DFS) and stop traversing down a path as soon as we find a value greater than the root's value. If we encounter a node with a value equal to the root's value, we continue traversing down that path.

Code Example (Python)

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right



def find_second_minimum_value(root: TreeNode) -> int:
    smallest = root.val
    second_smallest = float('inf')

    def traverse(node):
        nonlocal second_smallest
        if not node:
            return

        if smallest < node.val < second_smallest:
            second_smallest = node.val
        elif node.val == smallest:
            traverse(node.left)
            traverse(node.right)

    traverse(root)

    return second_smallest if second_smallest != float('inf') else -1

Big(O) Time Complexity Analysis

• Naive Approach: O(N), where N is the number of nodes in the tree. This is because we visit each node once to collect all unique values.
• Optimal Approach: O(N) in the worst-case scenario where all nodes have the same value as the root, forcing us to traverse the entire tree. However, in many cases, it will be less than O(N) because we stop traversing down paths once we find a value greater than the root.

Big(O) Space Complexity Analysis

• Naive Approach: O(N) in the worst-case scenario, where N is the number of nodes in the tree. This is because, in the worst case, all nodes have distinct values, and we store them in a set.
• Optimal Approach: O(H), where H is the height of the tree, due to the recursive call stack of the DFS. In the worst case (skewed tree), H = N, so it becomes O(N). In the average case (balanced tree), H = logN, so it becomes O(logN).

Edge Cases

1. Empty Tree: The problem statement mentions that the number of nodes is in the range [1, 1000], so an empty tree is not a valid input.
2. Tree with only one node: The function should return -1, as there is no second smallest value.
3. All nodes have the same value: The function should return -1, as there is no second smallest value.
4. Second smallest value is very large: The code should handle cases where the second smallest value is close to the maximum possible integer value.

Handling edge cases in code

The optimal solution gracefully handles the mentioned edge cases. If the tree contains only the smallest value, second_smallest remains float('inf'), and the function returns -1.