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# House Robber III LeetCode Solution

## Problem – House Robber III LeetCode Solution

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`

## House Robber III LeetCode Solution in C++

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int tryRob(TreeNode* root, int& l, int& r) {
if (!root)
return 0;

int ll = 0, lr = 0, rl = 0, rr = 0;
l = tryRob(root->left, ll, lr);
r = tryRob(root->right, rl, rr);

return max(root->val + ll + lr + rl + rr, l + r);
}

int rob(TreeNode* root) {
int l, r;
return tryRob(root, l, r);
}
};``````

## House Robber III LeetCode Solution in Java

``````public class Solution {
public int rob(TreeNode root) {
int[] num = dfs(root);
return Math.max(num, num);
}
private int[] dfs(TreeNode x) {
if (x == null) return new int;
int[] left = dfs(x.left);
int[] right = dfs(x.right);
int[] res = new int;
res = left + right + x.val;
res = Math.max(left, left) + Math.max(right, right);
return res;
}
}
``````

## House Robber III LeetCode Solution in Python

``````class Solution(object):
def rob(self, root):
"""
:type root: TreeNode
:rtype: int
"""
def superrob(node):
# returns tuple of size two (now, later)
# now: max money earned if input node is robbed
# later: max money earned if input node is not robbed

# base case
if not node: return (0, 0)

# get values
left, right = superrob(node.left), superrob(node.right)

# rob now
now = node.val + left + right

# rob later
later = max(left) + max(right)

return (now, later)

return max(superrob(root))
``````
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