Validate Binary Search Tree

mediumFREETreesBlind 75

Given the root of a binary tree (as a level-order array), determine if it is a valid binary search tree (BST).

A valid BST has:

Left subtree contains only nodes with keys less than the node's key
Right subtree contains only nodes with keys greater than the node's key
Both subtrees must also be valid BSTs

Example

Input

2,1,3

Output

True

Reasoning

The input array 2,1,3 represents a binary tree with root node 2, left child 1, and right child 3.
We check the left subtree: since 1 is less than 2, it satisfies the BST condition.
We check the right subtree: since 3 is greater than 2, it satisfies the BST condition.
Both subtrees are valid BSTs (as they are single nodes with no children), so the entire tree is a valid binary search tree, resulting in an output of True.

Constraints

1 <= number of nodes <= 10^4
-2^31 <= Node.val <= 2^31 - 1

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Concept primer

Background Knowledge

A binary search tree (BST) is a fundamental data structure in computer science, where each node has at most two children (i.e., left child and right child). The key characteristics of a BST are:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both subtrees must also be valid BSTs.

To understand this problem, it's essential to have a solid grasp of tree traversal techniques, such as in-order traversal, pre-order traversal, and post-order traversal. In-order traversal is particularly relevant, as it visits nodes in ascending order, which can help verify the BST property. Additionally, familiarity with recursive functions and base cases is crucial for solving this problem.

The concept of a valid BST is closely related to the idea of ordering and constraints. In a valid BST, each node's key is constrained by the keys of its ancestors, ensuring that the tree remains ordered. This constraint is what makes BSTs useful for efficient searching, inserting, and deleting nodes.

Algorithm/Approach

The general approach to solving this problem involves traversing the tree and checking the BST property at each node. This can be achieved through recursive or iterative methods. A common algorithm pattern is to use in-order traversal to visit nodes in ascending order and verify that each node's key is within the valid range.

Step-by-Step Strategy

To implement the solution:

Define a function to perform in-order traversal of the tree, which visits nodes in ascending order.
During traversal, check the BST property at each node by verifying that its key is within the valid range (i.e., greater than the previous node's key and less than the next node's key).
Use recursive functions or iterative methods to traverse the tree and check the BST property.
Handle base cases, such as an empty tree or a tree with a single node, which are inherently valid BSTs.

Common Pitfalls

When implementing the solution, watch out for:

Incorrectly handling edge cases, such as an empty tree or a tree with a single node.
Failing to properly update the valid range for each node during traversal.
Not correctly checking the BST property at each node, leading to false positives or false negatives.

Time & Space Complexity

The expected time complexity for this problem is $O(n)$ , where $n$ is the number of nodes in the tree, since we need to visit each node once during traversal. The space complexity depends on the implementation:

Recursive approaches typically have a space complexity of $O(h)$ , where $h$ is the height of the tree, due to the recursive call stack.
Iterative approaches can have a space complexity of $O(n)$ , depending on the data structures used to store nodes during traversal.

solution.py

Python 3.13

Test results