109. Convert Sorted List to Binary Search Tree
Given the head of a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example 1:
Input: head = [-10,-3,0,5,9] Output: [0,-3,9,-10,null,5] Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.
Example 2:
Input: head = [] Output: []
Example 3:
Input: head = [0] Output: [0]
Example 4:
Input: head = [1,3] Output: [3,1]
Constraints:
- The number of nodes in
headis in the range[0, 2 * 104]. -10^5 <= Node.val <= 10^5
Rust Solution
struct Solution;
use rustgym_util::*;
trait Convert {
fn convert(self, n: usize) -> (TreeLink, ListLink);
fn count(&self) -> usize;
}
impl Convert for ListLink {
fn count(&self) -> usize {
let mut p = self;
let mut n = 0;
while let Some(node) = p {
p = &node.next;
n += 1;
}
n
}
fn convert(self, n: usize) -> (TreeLink, ListLink) {
if n == 0 {
return (None, self);
}
if n == 1 {
let mut node = self.unwrap();
return (tree!(node.val), node.next.take());
}
let (left, next) = self.convert(n / 2);
let (middle, next) = next.convert(1);
let (right, next) = next.convert(n - n / 2 - 1);
let node = middle.unwrap();
node.borrow_mut().left = left;
node.borrow_mut().right = right;
(Some(node), next)
}
}
impl Solution {
fn sorted_list_to_bst(head: ListLink) -> TreeLink {
let n = head.count();
let (tree, _) = head.convert(n);
tree
}
}
#[test]
fn test() {
let head = list!(-10, -3, 0, 5, 9);
let res = tree!(0, tree!(-3, tree!(-10), None), tree!(9, tree!(5), None));
assert_eq!(Solution::sorted_list_to_bst(head), res);
}
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