285. Inorder Predecessor in BST §

Date: 2026-02-17 Difficulty: Medium Topics: BST, Binary Search Link: https://leetcode.com/problems/inorder-predecessor-in-bst/

Final Solution §

def inorderPredecessor(self, root: TreeNode, p: TreeNode) -> Optional[TreeNode]:
    predecessor = None
    while root:
        if root.val < p.val:
            predecessor = root
            root = root.right
        else:
            root = root.left
    return predecessor

Key Learnings §

• Exact mirror of inorder successor
• When going right (root.val < p.val), that node is a potential predecessor
• When root.val >= p.val, go left to find something smaller
• Got it right on the first attempt by applying the successor pattern

Nuances to Remember §

• O(H) time, O(1) space — same as successor
• Successor: update when going left; Predecessor: update when going right