LeetCode-in-Swift

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

To solve the “Validate Binary Search Tree” problem in Swift with the Solution class, follow these steps:

1. Define a method isValidBST in the Solution class that takes the root of a binary tree as input and returns true if the tree is a valid binary search tree (BST), and false otherwise.
2. Implement a recursive approach to validate if the given binary tree is a valid BST:
   • Define a helper method isValidBSTHelper that takes the root node, a lower bound, and an upper bound as input parameters.
   • In the isValidBSTHelper method, recursively traverse the binary tree nodes.
   • At each node, check if its value is within the specified bounds (lower bound and upper bound) for a valid BST.
   • If the node’s value violates the BST property, return false.
   • Otherwise, recursively validate the left and right subtrees by updating the bounds accordingly.
   • If both the left and right subtrees are valid BSTs, return true.
3. Call the isValidBSTHelper method with the root node and appropriate initial bounds to start the validation process.

Here’s the implementation of the isValidBST method in Swift:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init() { self.val = 0; self.left = nil; self.right = nil; }
 *     public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
 *     public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
 *         self.val = val
 *         self.left = left
 *         self.right = right
 *     }
 * }
 */
class Solution {
    func isValidBST(_ root: TreeNode?) -> Bool {
        return solve(root, Int.min, Int.max)
    }

    private func solve(_ root: TreeNode?, _ left: Int, _ right: Int) -> Bool {
        guard let root = root else {
            return true
        }
        if root.val <= left || root.val >= right {
            return false
        }
        return solve(root.left, left, root.val) && solve(root.right, root.val, right)
    }
}

This implementation recursively validates whether the given binary tree is a valid BST in O(n) time complexity, where n is the number of nodes in the tree.

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