借鑑DP思想: HouseRobberIII

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:

     3
    / \
   2   3
    \   \ 
     3   1

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

 

題目大意

給定一二叉樹，樹節點上有權值，從樹中選取一些不直接相鄰的節點，使得節點和最大

 

思考過程

1. 簡單粗暴，搜索完事，一個節點不外乎兩種狀況，選擇 or 不選擇；另外當前節點選擇以後，子節點不能被選擇；當前節點若不選擇，則又能夠分爲四種狀況

* 左選擇，右不選

* 右選，左不選

* 左右都選

* 左右都不選

2. 寫代碼，好的然而超時（固然-_-後來看了其餘的解答，才發現too young）

3. 由於看到有重複計算，因而朝動態規劃考慮，因而想出這麼個狀態

d[0][1]，其中0表示存儲遍歷到當前節點時，取當前節點能達到的最大值，而1則表示，不取當前節點能達到的最大值

又由於是樹節點，因此直接哈希表存儲d[TreeNode*][]

4. 遍歷順序呢，習慣性後序遍歷

5. 計算規則

// 顯然，當前節點取了，子節點不能取

d[TreeNode*][0] = TreeNodeVal + d[LeftChild][1] + d[RightChild][1] 

// 四種狀況

d[TreeNode*][1] = max(d[LeftChild][0] + d[RightChild][0], d[LeftChild][1] + d[RightChild][0], d[LeftChild][0] + d[RightChild][1], d[LeftChild][1] + d[RightChild][1]) 

 

6. 總算過了，附代碼；看了討論的思路以後，以爲真是too young，╮(╯▽╰)╭

 1 class Solution {
 2 public:
 3     int rob(TreeNode* root) {
 4         if (root == NULL) {
 5             return 0;
 6         }
 7         
 8         postOrder(root);
 9         return max(d[root][0], d[root][1]);
10     }
11 
12     void postOrder(TreeNode* itr) {
13         if (itr == NULL) {
14             return;
15         }
16 
17         postOrder(itr->left);
18         postOrder(itr->right);
19 
20         auto dItr = d.insert(pair<TreeNode*, vector<int>>(itr, vector<int>(2, 0)));
21         auto leftItr = dItr.first->first->left;
22         auto rightItr = dItr.first->first->right;
23 
24         int rL = dItr.first->first->left != NULL ? d[dItr.first->first->left][0] : 0;
25         int rR = dItr.first->first->right != NULL ? d[dItr.first->first->right][0] : 0;
26         int uL = dItr.first->first->left != NULL ? d[dItr.first->first->left][1] : 0;
27         int uR = dItr.first->first->right != NULL ? d[dItr.first->first->right][1] : 0;
28 
29         dItr.first->second[0] = dItr.first->first->val + uL + uR;
30         dItr.first->second[1] = max(max(max(rL + uR, uL + rR), rL + rR), uL + uR);
31     }
32 
33 private:
34     unordered_map<TreeNode*, vector<int>> d;
35 };

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