Return the root node of a binary search tree that matches the given preorder traversal.
(Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val. Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.)
It's guaranteed that for the given test cases there is always possible to find a binary search tree with the given requirements.
Example 1:
Input: [8,5,1,7,10,12] Output: [8,5,10,1,7,null,12]
Constraints:
1 <= preorder.length <= 1001 <= preorder[i] <= 10^8- The values of
preorderare distinct.
struct Solution;
use rustgym_util::*;
trait Postorder {
fn from_vec(preorder: &[i32], inorder: &[i32]) -> Self;
}
impl Postorder for TreeLink {
fn from_vec(preorder: &[i32], inorder: &[i32]) -> Self {
let n = preorder.len();
if n == 0 {
None
} else {
if n == 1 {
tree!(preorder[0])
} else {
let i = inorder.binary_search(&preorder[0]).unwrap();
tree!(
preorder[0],
TreeLink::from_vec(&preorder[1..=i], &inorder[0..i]),
TreeLink::from_vec(&preorder[i + 1..], &inorder[i + 1..])
)
}
}
}
}
impl Solution {
fn bst_from_preorder(preorder: Vec<i32>) -> TreeLink {
let mut inorder: Vec<i32> = preorder.clone();
inorder.sort_unstable();
TreeLink::from_vec(&preorder, &inorder)
}
}
#[test]
fn test() {
let preorder = vec![8, 5, 1, 7, 10, 12];
let res = tree!(8, tree!(5, tree!(1), tree!(7)), tree!(10, None, tree!(12)));
assert_eq!(Solution::bst_from_preorder(preorder), res);
}