Move from brute-force thinking to an efficient approach using dynamic programming strategy.
The thief has found himself a new place for his thievery again. There is only one entrance to this area, called 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 form a binary tree. It will automatically contact the police if two directly-linked houses were broken into on the same night.
Given the root of the binary tree, return the maximum amount of money the thief can rob without alerting the police.
Example 1:
Input: root = [3,2,3,null,3,null,1] Output: 7 Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
Input: root = [3,4,5,1,3,null,1] Output: 9 Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.
Constraints:
[1, 104].0 <= Node.val <= 104Problem summary: The thief has found himself a new place for his thievery again. There is only one entrance to this area, called 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 form a binary tree. It will automatically contact the police if two directly-linked houses were broken into on the same night. Given the root of the binary tree, return the maximum amount of money the thief can rob without alerting the police.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming · Tree
[3,2,3,null,3,null,1]
[3,4,5,1,3,null,1]
house-robber)house-robber-ii)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #337: House Robber III
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int rob(TreeNode root) {
int[] ans = dfs(root);
return Math.max(ans[0], ans[1]);
}
private int[] dfs(TreeNode root) {
if (root == null) {
return new int[2];
}
int[] l = dfs(root.left);
int[] r = dfs(root.right);
return new int[] {root.val + l[1] + r[1], Math.max(l[0], l[1]) + Math.max(r[0], r[1])};
}
}
// Accepted solution for LeetCode #337: House Robber III
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func rob(root *TreeNode) int {
var dfs func(*TreeNode) (int, int)
dfs = func(root *TreeNode) (int, int) {
if root == nil {
return 0, 0
}
la, lb := dfs(root.Left)
ra, rb := dfs(root.Right)
return root.Val + lb + rb, max(la, lb) + max(ra, rb)
}
a, b := dfs(root)
return max(a, b)
}
# Accepted solution for LeetCode #337: House Robber III
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def rob(self, root: Optional[TreeNode]) -> int:
def dfs(root: Optional[TreeNode]) -> (int, int):
if root is None:
return 0, 0
la, lb = dfs(root.left)
ra, rb = dfs(root.right)
return root.val + lb + rb, max(la, lb) + max(ra, rb)
return max(dfs(root))
// Accepted solution for LeetCode #337: House Robber III
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);
}
// Accepted solution for LeetCode #337: House Robber III
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function rob(root: TreeNode | null): number {
const dfs = (root: TreeNode | null): [number, number] => {
if (!root) {
return [0, 0];
}
const [la, lb] = dfs(root.left);
const [ra, rb] = dfs(root.right);
return [root.val + lb + rb, Math.max(la, lb) + Math.max(ra, rb)];
};
return Math.max(...dfs(root));
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.