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);
}
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