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.

Analysis

前两题 I, II的变形升级，放置在Binary Tree中，但实际上还是一个DP问题。

LeetCode的这个讨论很详尽：https://leetcode.com/problems/house-robber-iii/discuss/79330/Step-by-step-tackling-of-the-problem?orderBy=most_votes

the two conditions for dynamic programming: "optimal substructure" + "overlapping of subproblems", we actually have a DP problem

递推关系在于是否取root，则对应left，right子树分别取或者不取。

Solution

DP with Memorization O(n) space cost (n is the total number of nodes; stack cost for recursion is not counted).

public int rob(TreeNode root) {
    return robSub(root, new HashMap<>());
}

private int robSub(TreeNode root, Map<TreeNode, Integer> map) {
    if (root == null) return 0;
    if (map.containsKey(root)) return map.get(root);

    int val = 0;

    if (root.left != null) {
        val += robSub(root.left.left, map) + robSub(root.left.right, map);
    }

    if (root.right != null) {
        val += robSub(root.right.left, map) + robSub(root.right.right, map);
    }

    val = Math.max(val + root.val, robSub(root.left, map) + robSub(root.right, map));
    map.put(root, val);

    return val;
}

DP (0ms 100% AC)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int rob(TreeNode root) {
        int[] res = robHelper(root);
        return Math.max(res[0], res[1]);
    }
    int[] robHelper(TreeNode root) {
        if (root == null) return new int[2];
        int[] left = robHelper(root.left);
        int[] right = robHelper(root.right);
        int[] res = new int[2];
        res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        res[1] = root.val + left[0] + right[0];
        return res;
    }
}