LeetCode 第47天 | 打家劫舍I 打家劫舍II 打家劫舍III 动态规划 首次遇见树形dp

198. 打家劫舍

class Solution {
public:
    //这就是大名鼎鼎的打家劫舍吗
    int rob(vector& nums) {
        if (nums.size() == 0) return 0;
        if (nums.size() == 1) return nums[0];
        vector dp(nums.size(),0);
        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);
        for (int i = 2; i  
213. 打家劫舍 II
 
class Solution {
public:
    // 首尾只能选一个较大的
    int rob(vector& nums) {
        if (nums.size() == 0) return 0;
        if (nums.size() == 1) return nums[0];
        int res1 = rangeRob(nums, 0, nums.size()-1);
        int res2 = rangeRob(nums, 1, nums.size());
        return max(res1, res2);
    }
    int rangeRob(vector& nums, int start, int end) {
        if (start == end-1 ) return nums[start];
        vector dp(nums.size(), 0);
        dp[start] = nums[start];
        dp[start+1] = max(nums[start], nums[start+1]);
        for (int i = start+2; i  
337. 打家劫舍 III
 
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int rob(TreeNode* root) {
        vector res = robTree(root);
        return max(res[0], res[1]);
    }
    vector robTree(TreeNode* root) {
        if (root == nullptr) return vector{0,0};
        vector left = robTree(root->left);
        vector right = robTree(root->right);
        int val1 = root->val + left[0] + right[0];
        int val2 = max(left[0], left[1]) + max(right[0], right[1]);
        return {val2, val1};
    }
};