RUST GYM v0.2.11

98. Validate Binary Search Tree

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

Rust Solution

struct Solution;
use rustgym_util::*;

trait Inorder {
    fn inorder(&self, visit: &mut dyn FnMut(i32));
}

impl Inorder for TreeLink {
    fn inorder(&self, visit: &mut dyn FnMut(i32)) {
        if let Some(node) = self {
            let node = node.borrow();
            Self::inorder(&node.left, visit);
            visit(node.val);
            Self::inorder(&node.right, visit);
        }
    }
}

impl Solution {
    fn is_valid_bst(root: TreeLink) -> bool {
        let mut prev: Option<i32> = None;
        let mut res = true;
        root.inorder(&mut |x| {
            if let Some(y) = prev {
                if x <= y {
                    res = false;
                }
            }
            prev = Some(x);
        });
        res
    }
}

#[test]
fn test() {
    let root = tree!(2, tree!(1), tree!(3));
    assert_eq!(Solution::is_valid_bst(root), true);
    let root = tree!(5, tree!(1), tree!(4, tree!(3), tree!(6)));
    assert_eq!(Solution::is_valid_bst(root), false);
}

Having problems with this solution? Click here to submit an issue on github.