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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