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.
题目大意
给定一二叉树，树节点上有权值，从树中选取一些不直接相邻的节点，使得节点和最大
思考过程
1. 简单粗暴，搜索完事，一个节点不外乎两种情况，选择 or 不选择；另外当前节点选择之后，子节点不能被选择；当前节点若不选择，则又可以分为四种情况
* 左选择，右不选
* 右选，左不选
* 左右都选
* 左右都不选
2. 写代码，好的然而超时（当然-_-后来看了其他的解答，才发现too young）
3. 因为看到有重复计算，于是朝动态规划考虑，于是想出这么个状态
d[0][1]，其中0表示存储遍历到当前节点时，取当前节点能达到的最大值，而1则表示，不取当前节点能达到的最大值
又因为是树节点，所以直接哈希表存储d[TreeNode*][]
4. 遍历顺序呢，习惯性后序遍历
5. 计算规则
// 显然，当前节点取了，子节点不能取
d[TreeNode*][0] = TreeNodeVal + d[LeftChild][1] + d[RightChild][1]
// 四种情况
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. 总算过了，附代码；看了讨论的思路之后，觉得真是too young，╮(╯▽╰)╭
class Solution {
public:
int rob(TreeNode* root) {
if (root == NULL) {
return ;
}
postOrder(root);
return max(d[root][], d[root][]);
}
void postOrder(TreeNode* itr) {
if (itr == NULL) {
return;
}
postOrder(itr->left);
postOrder(itr->right);
auto dItr = d.insert(pair<TreeNode*, vector<int>>(itr, vector<int>(, )));
auto leftItr = dItr.first->first->left;
auto rightItr = dItr.first->first->right;
int rL = dItr.first->first->left != NULL ? d[dItr.first->first->left][] : ;
int rR = dItr.first->first->right != NULL ? d[dItr.first->first->right][] : ;
int uL = dItr.first->first->left != NULL ? d[dItr.first->first->left][] : ;
int uR = dItr.first->first->right != NULL ? d[dItr.first->first->right][] : ;
dItr.first->second[] = dItr.first->first->val + uL + uR;
dItr.first->second[] = max(max(max(rL + uR, uL + rR), rL + rR), uL + uR);
}
private:
unordered_map<TreeNode*, vector<int>> d;
};