LeetCode-in-Rust
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
impl Solution {
pub fn is_interleave(s1: String, s2: String, s3: String) -> bool {
let mut state = vec![vec![None::<bool>; s2.len() + 1]; s1.len() + 1];
is_interleave_0(&s1, &s2, &s3, &mut state)
}
}
fn is_interleave_0(s1: &str, s2: &str, s3: &str, state: &mut Vec<Vec<Option<bool>>>) -> bool {
if let Some(result) = state[s1.len()][s2.len()] {
return result;
}
if s1.len() + s2.len() != s3.len() {
return false;
}
if s1.is_empty() {
return s2 == s3;
}
if s2.is_empty() {
return s1 == s3;
}
let (head1, tail1) = s1.split_at(1);
let (head2, tail2) = s2.split_at(1);
let (head3, tail3) = s3.split_at(1);
if head1 == head3 && is_interleave_0(tail1, s2, tail3, state) {
state[s1.len()][s2.len()] = Some(true);
return true;
} else if head2 == head3 && is_interleave_0(s1, tail2, tail3, state) {
state[s1.len()][s2.len()] = Some(true);
return true;
} else {
state[s1.len()][s2.len()] = Some(false);
return false;
}
}