Leetcode 894. All Possible Full Binary Trees

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[@probabilitycodingisfunis1]

unordered_mapdp;
vector allPossibleFBT(int n) {
if(n%2==0)return {};
if(n==1)return {new TreeNode()};
if(dp.find(n)!=dp.end())return dp[n];
vectorans;
for(int i=1;ileft = l;
node->right = r;
ans.push_back(node);
}
}
}
return dp[n] = ans;
}

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Rust implementation ->
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option,
// pub right: Option,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::{rc::Rc, cell::RefCell, collections::HashMap};
type T = Option;
macro_rules! treeNode {
($n:expr) => { Some(Rc::new(RefCell::new(TreeNode::new($n)))) };
($a:expr, $b:expr) => { Some(Rc::new(RefCell::new(TreeNode { val: 0, left: $a, right: $b, })))};
}
impl Solution {
pub fn all_possible_fbt(n: i32) -> Vec {
fn bk(n: i32, dp: &mut HashMap) -> Rc {
match n {
0 => dp.get(&0).unwrap().clone(),
1 => dp.get(&1).unwrap().clone(),
_ if dp.contains_key(&n) => dp.get(&n).unwrap().clone(),
_ => {
let mut res = Vec::new();
for l in (1..n).step_by(2) {
let r = n - 1 - l;
for t1 in bk(l, dp).iter() {
for t2 in bk(r, dp).iter() { res.push(treeNode!(t1.clone(), t2.clone())); }
}
}
let res = Rc::new(res);
dp.insert(n, res.clone());
res
}
}
}
let result = bk( n, &mut HashMap::from([ (0, Rc::new(Vec::::new())), (1, Rc::new(vec![treeNode!(0)])), ]));
Rc::try_unwrap(result).unwrap()
}
}