?? Unlocking the Univalued Binary Tree Mystery! ??

?? Problem: 965 Univalued Binary Tree

A binary tree is uni-valued if every node in the tree has the same value. Given the root of a binary tree, return true if the given tree is uni-valued, or false otherwise.

?? Approach - Recursive DSF preorder:

The algorithm recursively traverses the binary tree, comparing each node's value with the target value obtained from the root node. It returns True in the base case when the current node is None, indicating that an empty subtree is univalued. If a node's value differs from the target value, it returns False. Otherwise, it recursively checks the left and right subtrees, returning True only if both are univalued.

?? Complexity Analysis:

Time complexity: O(n), where n represents the number of nodes in the binary tree. The algorithm traverses each node exactly once during the recursive exploration.

Space complexity: O(h), where h signifies the height of the binary tree. The space required for the recursive function calls is proportional to the height of the tree.

?? Solution Links:

Recursive DFS preorder: https://lnkd.in/dAHUhTRJ

Iterative DFS inorder: https://lnkd.in/dhCTHrsJ

?? Key Takeaways:

• In the recursive exploration, the base case returns True when the current node is None, indicating that an empty subtree is univalued.
• This strategic use of True in the base case ensures that the algorithm correctly handles empty subtrees and contributes to the overall efficiency of the solution.

?? Join the Discussion:

I'm thrilled to engage with fellow professionals in exploring the intricacies of binary tree traversal and problem-solving strategies. Have you encountered challenges in handling univalued binary trees? Let's exchange insights, experiences, and best practices to deepen our understanding of algorithms and data structures! Connect with me on LinkedIn to continue the discussion.

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