House Robber III

House Robber III - Problem

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.

Input & Output

Example 1 — Basic Tree

$ Input: root = [3,2,3,null,3,null,1]

› Output: 7

💡 Note: Rob nodes 3, 3, and 1 for maximum sum. Cannot rob adjacent nodes (parent-child), so skip the middle level nodes with value 2 and 3.

Example 2 — Simple Tree

$ Input: root = [3,4,5,1,3,null,1]

› Output: 9

💡 Note: Rob nodes 4, 3, and 1 for sum = 4 + 3 + 1 = 8. Or rob nodes 3 and 5 for sum = 3 + 5 = 8. Or rob root 3 and leaves 1,3,1 for sum = 3 + 1 + 3 + 1 = 8. Actually, rob 4 + 5 = 9 is optimal.

Example 3 — Single Node

$ Input: root = [5]

› Output: 5

💡 Note: Only one house to rob, so rob it for value 5.

Constraints

The number of nodes in the tree is in the range [1, 10⁴]
0 ≤ Node.val ≤ 10⁴

Visualization

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Asked in

a Amazon 15 f Facebook 12 G Google 8 M Microsoft 6

The key insight is that for each node we have two choices: rob it (and skip children) or skip it (and potentially rob children). The optimal solution uses DFS with state tracking, maintaining [rob, skip] values for each subtree. Time: O(n), Space: O(h) where h is tree height.

Common Approaches

✓ Hash Map

⏱️ Time: N/A Space: N/A

Brute Force Recursion

⏱️ Time: O(2ⁿ) Space: O(h)

For each node, calculate two scenarios: robbing it (skip children) or not robbing it (can rob children). Take the maximum of both options.

Memoization (Top-Down DP)

⏱️ Time: O(n) Space: O(n)

Same recursive logic as brute force, but store results in a hash map to avoid recalculating the same subtree multiple times.

Optimal DFS (Bottom-Up State Tracking)

⏱️ Time: O(n) Space: O(h)

For each node, return a pair [rob_max, skip_max] representing maximum money when robbing or skipping that node. Build answer bottom-up.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Smart Actions

💡 Explanation

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