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BST In order Predecessor
Unit 8 Session 2 (Click for link to problem statements)
- 💡 Difficulty: Medium
- ⏰ Time to complete: 20 mins
- 🛠️ Topics: Trees, Binary Search Trees, In-Order Predecessor
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: What should be done if there is no in-order predecessor available?
- Answer: Return None if there is no in-order predecessor, such as when the node is the leftmost node.
HAPPY CASE
Input: BST with nodes [20, 10, 30, 5, 15], search in-order predecessor of 10
Output: 5
Explanation: 5 is the in-order predecessor of 10 as it is the largest value less than 10.
EDGE CASE
Input: BST with nodes [20, 10, 30, 5, 15], search in-order predecessor of 5
Output: None
Explanation: 5 is the leftmost node, so it has no predecessor.
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 problem involves navigating through a binary search tree (BST) to find the in-order predecessor, utilizing the properties of BSTs to efficiently find the required node.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Utilize the properties of BSTs to find the in-order predecessor of a node.
1) If the node has a left subtree, the in-order predecessor is the rightmost node of this subtree.
2) If the node has no left subtree, the predecessor is one of its ancestors; specifically, the nearest ancestor for which this node is in the right subtree.
3) Traverse the tree from the root to find the node and keep track of the potential predecessor.
- Incorrectly identifying the in-order predecessor or failing to handle the absence of a left subtree.
Implement the code to solve the algorithm.
def inorder_predecessor(root, current):
"
Find the in-order predecessor of a given node in a BST.
"
predecessor = None
while root:
if current.val > root.val:
predecessor = root
root = root.right
elif current.val < root.val:
root = root.left
else:
if root.left:
predecessor = find_rightmost(root.left)
break
return predecessor
def find_rightmost(node):
"
Find the rightmost (largest) node starting from the given node.
"
while node.right:
node = node.right
return node
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
- Ensure correct in-order predecessor identification in various configurations, especially where nodes do not have a left subtree.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
-
Time Complexity:
O(h)
where h is the height of the tree. This ensures efficient traversal up to the root or down to a leaf, depending on the node's position. -
Space Complexity:
O(1)
as it only requires maintaining a few pointers without additional structures.