337. House Robber III
The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.
Determine the maximum amount of money the thief can rob tonight without alerting the police.
Example 1:
Input: [3,2,3,null,3,null,1]
3
/ \
2 3
\ \
3 1
Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
Input: [3,4,5,1,3,null,1]
3
/ \
4 5
/ \ \
1 3 1
Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.
Rust Solution
struct Solution;
use rustgym_util::*;
trait Postorder {
fn postorder(&self) -> (i32, i32);
}
impl Postorder for TreeLink {
fn postorder(&self) -> (i32, i32) {
if let Some(node) = self {
let node = node.borrow();
let val = node.val;
let left = node.left.postorder();
let right = node.right.postorder();
(
val + left.1 + right.1,
left.0.max(left.1) + right.0.max(right.1),
)
} else {
(0, 0)
}
}
}
impl Solution {
fn rob(root: TreeLink) -> i32 {
let (a, b) = root.postorder();
a.max(b)
}
}
#[test]
fn test() {
let root = tree!(3, tree!(2, None, tree!(3)), tree!(3, None, tree!(1)));
let res = 7;
assert_eq!(Solution::rob(root), res);
let root = tree!(3, tree!(4, tree!(1), tree!(3)), tree!(5, None, tree!(1)));
let res = 9;
assert_eq!(Solution::rob(root), res);
}
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