1. Description

House Robber III

2. Solution

Version 1

/**
 * 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) {
        if(!root) {
            return 0;
        }
        int sum = 0;
        if(root->left) {
            sum = sum + rob(root->left->left) + rob(root->left->right);
        }
        if(root->right) {
            sum = sum + rob(root->right->left) + rob(root->right->right);
        }
        return max(root->val + sum, rob(root->left) + rob(root->right));
    }
};

Version 2

/**
 * 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) {
        unordered_map<TreeNode*, int> m;
        return rob(root, m);
    }

private:
    int rob(TreeNode* root, unordered_map<TreeNode*, int>& m) {
        if(!root) {
            return 0;
        }
        if(m.find(root) != m.end()) {
            return m[root];
        }
        int sum = 0;
        if(root->left) {
            sum = sum + rob(root->left->left, m) + rob(root->left->right, m);
        }
        if(root->right) {
            sum = sum + rob(root->right->left, m) + rob(root->right->right, m);
        }
        m[root] = max(root->val + sum, rob(root->left, m) + rob(root->right, m));
        return m[root];
    }
};