选择二叉树中不相邻的节点，使得节点之和为最大。

/// Memory Search
/// Time Complexity: O(n), where n is the nodes' number in the tree
/// Space Complexity: O(n)
class Solution {

public:
    int rob(TreeNode* root) {
        return rob(root, true);
    }

private:
    int rob(TreeNode* root, bool include){

        if(root == NULL)
            return 0;

        int res = rob(root->left, true) + rob(root->right, true);
        if(include)
            res = max(root->val + rob(root->left, false) + rob(root->right, false),
                      res);

        return res;
    }
};
/// Redefine the recursive function and return a two-element array
/// represent the include and exclude maximum result for robbing the node
///
/// Time Complexity: O(n), where n is the nodes' number in the tree
/// Space Complexity: O(1)
class Solution {

public:
    int rob(TreeNode* root) {
        vector<int> result = tryRob(root);
        return max(result[0], result[1]);
    }

private:
    vector<int> tryRob(TreeNode* root){

        if(root == NULL)
            return vector<int>(2, 0);

        vector<int> resultL = tryRob(root->left);
        vector<int> resultR = tryRob(root->right);
        vector<int> res(2, 0);

        res[0] = resultL[1] + resultR[1];
        res[1] = max(res[0], root->val + resultL[0] + resultR[0]);

        return res;
    }
};