Algorithm
Problem Name: 337. 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.
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, 104]. 0 <= Node.val <= 104
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int rob(struct TreeNode* root) {
int val, val_next;
if (!root) return 0;
val = root->val;
if (root->left) val += rob(root->left->left) + rob(root->left->right);
if (root->right) val += rob(root->right->left) + rob(root->right->right);
val_next = rob(root->left) + rob(root->right);
return val > val_next ? val : val_next;
}
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int rob(TreeNode root) {
int[] ans = helper(root);
return Math.max(ans[0], ans[1]);
}
private int[] helper(TreeNode root) {
if (root == null) {
return new int[]{0, 0};
}
int[] left = helper(root.left);
int[] right = helper(root.right);
int robbed = root.val + left[1] + right[1];
int notRobbed = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
return new int[]{robbed, notRobbed};
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const rob = function(root) {
return Math.max(...dfs(root))
}
function dfs(node) {
if (node == null) return [0, 0]
const left = dfs(node.left)
const right = dfs(node.right)
return [
node.val + left[1] + right[1],
Math.max(left[0], left[1]) + Math.max(right[0], right[1])
]
}
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def rob(self, root):
def dfs(node):
if not node: return 0, 0
l, r = dfs(node.left), dfs(node.right)
return max(l) + max(r), node.val + l[0] + r[0]
return max(dfs(root))
Input
Output