House Robber III - Medium - L - Star
Published:
Question
- https://leetcode.com/problems/house-robber-iii/
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.
Approach
- Recursive
- DP
Solution
class Solution {
public int rob(TreeNode root) {
if (root == null) {
return 0;
}
// reform tree into array-based tree
ArrayList<Integer> tree = new ArrayList<>();
HashMap<Integer, ArrayList<Integer>> graph = new HashMap<>();
graph.put(-1, new ArrayList<>());
int index = -1;
// we use two Queue to store node and index
Queue<TreeNode> q_node = new LinkedList<>();
q_node.add(root);
Queue<Integer> q_index = new LinkedList<>();
q_index.add(index);
while (q_node.size() > 0) {
TreeNode node = q_node.poll();
int parentIndex = q_index.poll();
if (node != null) {
index++;
tree.add(node.val);
graph.put(index, new ArrayList<>());
graph.get(parentIndex).add(index);
// push new node into Queue
q_node.add(node.left);
q_index.add(index);
q_node.add(node.right);
q_index.add(index);
}
}
// represent the maximum start by node i with robbing i
int[] dpRob = new int[index + 1];
// represent the maximum start by node i without robbing i
int[] dpNotRob = new int[index + 1];
for (int i = index; i >= 0; i--) {
ArrayList<Integer> children = graph.get(i);
if (children == null || children.size() == 0) {
// if is leaf
dpRob[i] = tree.get(i);
dpNotRob[i] = 0;
} else {
dpRob[i] = tree.get(i);
for (int child : children) {
dpRob[i] += dpNotRob[child];
dpNotRob[i] += Math.max(dpRob[child], dpNotRob[child]);
}
}
}
return Math.max(dpRob[0], dpNotRob[0]);
}
}