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***.
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:
The number of nodes in the tree is in the range
[1, 10^4].0 <= Node.val <= 10^4
Solution
/**
* 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[] out = robRec(root);
return Math.max(out[0], out[1]);
}
private int[] robRec(TreeNode curr) {
if (curr == null) {
return new int[] {0, 0};
}
int[] left = robRec(curr.left);
int[] right = robRec(curr.right);
int[] out = new int[2];
// 1. If choosing to take the house
out[0] = curr.val + left[1] + right[1];
// 2. If choosing to skip the house
out[1] = left[0] + right[0];
// 3. Best Solution at house
out[0] = Math.max(out[0], out[1]);
return out;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).