LeetCode-in-All

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;
    }
}

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