LeetCode in Elixir

98. Validate Binary Search Tree

Medium

Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid BST is defined as follows:

• 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 the left and right subtrees must also be binary search trees.

Example 1:

Input: root = [2,1,3]

Output: true

Example 2:

Input: root = [5,1,4,null,null,3,6]

Output: false

Explanation: The root node’s value is 5 but its right child’s value is 4.

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -231 <= Node.val <= 231 - 1

Solution

# Definition for a binary tree node.
#
# defmodule TreeNode do
#   @type t :: %__MODULE__{
#           val: integer,
#           left: TreeNode.t() | nil,
#           right: TreeNode.t() | nil
#         }
#   defstruct val: 0, left: nil, right: nil
# end

defmodule Solution do
  @spec is_valid_bst(root :: TreeNode.t() | nil) :: boolean
  def is_valid_bst(root) do
    root |> inorder_check() |> is_list()
  end

  @spec inorder_check(root :: TreeNode.t() | nil) :: [integer] | :invalid
  def inorder_check(root) do
    inorder_check(root, [])
  end

  def inorder_check(nil, acc), do: acc

  def inorder_check(root, acc) do
    # root comes in between in inorder traversal
    case inorder_check(root.left, acc) do
      [x | _] = acc ->
        if x >= root.val do
          :invalid
        else
          acc = [root.val | acc]
          inorder_check(root.right, acc)
        end

      [] ->
        acc = [root.val | acc]
        inorder_check(root.right, acc)

      _ ->
        :invalid
    end
  end
end

This site is open source. Improve this page.