Ceil in BST
A medium BST problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- BST
- Sheets
- 1
- Core for
- 2 roles
- Platform
- GeeksforGeeks
The problem
Given the root of a binary search tree and an integer x, find the ceiling of x. The ceiling is the smallest value in the BST that is greater than or equal to x. If no such value exists, return -1.
Example 1
- Input
- root=[6,2,8,0,4,7,9,null,null,3,5], x=5
- Output
- 5
- Why
- Value 5 exists in the tree, so the ceiling of 5 is 5.
Example 2
- Input
- root=[6,2,8,0,4,7,9,null,null,3,5], x=11
- Output
- -1
- Why
- No value in the tree is >= 11.
Example 3
- Input
- root=[1,0,2], x=2
- Output
- 2
- Why
- Value 2 exists in the tree.
Constraints
- The number of nodes is in the range [1, 104].
- -109 <= Node.val <= 109
- -109 <= x <= 109
How to think about it
Updated 2026-09-09Ceiling is the exact dual of floor. When the current value is at least x, it qualifies as a candidate ceiling, but smaller candidates might still hide in its left subtree. When the current value is strictly less than x, neither it nor its left descendants can qualify, forcing the search to the right.
Approaches, worst first
Full inorder search
time O(n) · space O(h)
Walk through nodes in ascending order until hitting the first value >= x. Correct and stops early when ceiling exists early, but still takes O(n) in the worst case when the target is larger than all elements.
Iterative candidate trackingWrite this one
time O(h) · space O(1)
Initialize `ceil = -1`. Walk down the tree: if `node.val == x`, return x. If `node.val > x`, record `ceil = node.val` and branch left to check if an even tighter ceiling exists. If `node.val < x`, branch right without updating `ceil`.
Where people lose marks · 2
- Returning prematurely on `node.val > x`. That node is a candidate ceiling, not necessarily the tightest one; the search must continue down `node.left`.
- Incorrect sentinel return: if all values in the tree are strictly smaller than x, ensure the function returns -1 as specified rather than crashing or returning an uninitialized variable.
The theory behind it
BST — the ground this problem stands on. All BST problems
What BST is
A binary search tree is a binary tree that enforces a strict ordering rule at every node. Every value stored in a node's left branch must be smaller than the node itself, and every value in its right branch must be larger. Because this rule holds true everywhere down the tree, searching for a value does not require checking every branch. At each step, a single comparison lets you discard an entire half of the remaining nodes.
When to reach for it
Reach for a binary search tree when problems mention sorted tree structures, finding the kth smallest element, checking tree validity, or searching within a range of values. Signals include queries for the next greater element in a dynamic set, finding the lowest common ancestor in a sorted tree, or converting sorted arrays into height-balanced trees. When data must remain sorted while supporting continuous insertions and lookups, BST properties are the target.
How the pattern works
Use the ordering property to prune entire subtrees. When looking for a key, move left if the target is smaller than the current node, or move right if it is larger, stopping when values match or when reaching a null link. For validating a tree, do not merely compare a node to its immediate children; pass valid lower and upper value bounds down through recursive calls. Walking a valid binary search tree in-order visits all values in strictly increasing numerical order.
What each operation costs
| Operation | Time |
|---|---|
| search, insert, or delete on balanced tree | O(log n) |
| search, insert, or delete on degenerate line tree | O(n) |
| in-order traversal visiting all nodes in sorted order | O(n) |
What usually goes wrong with BST
- Validating a tree by checking only immediate children against the parent, missing violations where a node deep in the left subtree is larger than an ancestor higher up.
- Allowing duplicate values on the wrong side when the problem specification requires strictly smaller values on the left and strictly greater on the right.
- Deleting a node with two children by removing it directly instead of replacing its value with its in-order predecessor or successor and deleting that leaf instead.
Which roles need this problem
BST is a core topic for these 2 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 6 more roles, including Full-Stack Developer, Android Developer, iOS Developer.
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