Binary Tree Is Leaf

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Unit 8 Session 1 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 10 mins
• 🛠️ Topics: Trees, Binary Trees, Search Algorithms

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• Question: How to handle a tree with duplicate values?
  • Answer: The problem states that all nodes have unique values, so this scenario does not apply.

HAPPY CASE
Input: TreeNode(10, TreeNode(5, TreeNode(2), TreeNode(7)), TreeNode(15)), value = 7
Output: True
Explanation: The node with value 7 is a leaf node (nodes 2 and 7 are leaves, 7 is checked).

EDGE CASE
Input: TreeNode(10, TreeNode(5), None), value = 5
Output: False
Explanation: Node 5 is not a leaf because it has a child.

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

This is a targeted search problem where the goal is to identify whether a node with a given value is a leaf in the tree. It involves a traversal with a condition check.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the tree to find a node with the specified value and check if it is a leaf (no children).

1) If the tree is empty (root is None), return False.
2) If the current node's value matches the target, check if it has no children.
3) If it matches and is a leaf, return True.
4) Otherwise, recursively search in the left and right subtrees.

⚠️ Common Mistakes

• Incorrectly identifying a node as a leaf when it has one or more children.

4: I-mplement

Implement the code to solve the algorithm.

def is_leaf(root, value):
    "
    Search the entire binary tree rooted at `root` for a node with the specified `value`
    and check if it is a leaf. A leaf node is a node that has no children.
    "
    if root is None:
        return False  # Node not found, so it's not a leaf

    # If the current node's value matches the search value, check if it's a leaf
    if root.val == value:
        return root.left is None and root.right is None

    # Search in the left and right subtrees
    return is_leaf(root.left, value) or is_leaf(root.right, value)

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Test with trees where the target node varies in position (root, middle, leaf) to ensure robustness in identifying leaf status.

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

• Time Complexity: O(n) where n is the number of nodes in the tree, as the worst case requires visiting every node.
• Space Complexity: O(h) where h is the height of the tree, due to recursion. This could be O(n) in a skewed tree.