House Robber III

Desicription

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the “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 forms a binary tree”. It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

Example 1:

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Input: [3,2,3,null,3,null,1]

3
/ \
2 3
\ \
3 1

Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

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Input: [3,4,5,1,3,null,1]

3
/ \
4 5
/ \ \
1 3 1

Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

Solution

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/**
* 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 rob(TreeNode* root) {
std::function<int(TreeNode*, int&, int&)> dfs = [&](TreeNode* root, int& left, int& right) {
if(root == nullptr) {
return 0;
}

int leftLeft = 0;
int leftRight = 0;
int rightLeft = 0;
int rightRight = 0;

left = dfs(root->left, leftLeft, leftRight);
right = dfs(root->right, rightLeft, rightRight);

return std::max(root->val + leftLeft + leftRight + rightLeft + rightRight, left + right);
};

int left = 0;
int right = 0;
return dfs(root, left, right);
}
};