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.
Test Cases
Input:
(TreeNode) root = [3,2,3,null,3,null,1]
Output:
(int) 7
Explanation:
3
/ \
2 3
\ \
3 1
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Solution
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int rob(TreeNode* root) {
vector<int> res = robSub(root);
return max(res[0], res[1]);
}
vector<int> robSub(TreeNode* root) {
vector<int> res(2, 0);
if (root == nullptr) {
return res;
}
vector<int> left = robSub(root->left);
vector<int> right = robSub(root->right);
res[0] = max(left[0], left[1]) + max(right[0], right[1]);
res[1] = root->val + left[0] + right[0];
return res;
}
};/**
* 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[] res = robSub(root);
return Math.max(res[0], res[1]);
}
private int[] robSub(TreeNode root) {
if (root == null) return new int[2];
int[] left = robSub(root.left);
int[] right = robSub(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;
}
}from typing import Optional
# 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:
res = self.robsub(root)
return max(res[0], res[1])
def robsub(self, root):
if not root:
return [0, 0]
left = self.robsub(root.left)
right = self.robsub(root.right)
res = [0, 0]
res[0] = max(left[0], left[1]) + max(right[0], right[1])
res[1] = root.val + left[0] + right[0]
return res