[LeetCode] House Robber III 打家劫舍之三

1. 比较有意思的一道题目。当抢劫一个节点的时候，它的左右子节点肯定不可以抢。当不抢劫一个节点的时候，它的左右子节点可抢，可不抢。我们在处理的过程中取最大值即可。
   #include
   #include
   #include
   #include
   #include
   using namespace std;
   struct TreeNode
   {
   int val;
   TreeNode lChild;
   TreeNode rChild;
   TreeNode(int val, TreeNodel = nullptr, TreeNoder = nullptr)
   {
   this->val = val;
   this->lChild = l;
   this->rChild = r;
   }
   };
   TreeNode *create(int intarr[], int length, int i)
   {
   TreeNode *node = nullptr;

   if (i < length && intarr[i] != -1)
   {
   node = new TreeNode(intarr[i]);
   node->lChild = create(intarr, length, i * 2 + 1);
   node->rChild = create(intarr, length, i * 2 + 2);
   }
   return node;
   }
   int maxT(int a, int b)
   {
   if (a > b)
   {
   return a;
   }
   return b;
   }
   int robberNode(TreeNode *node);
   int NotRobberNode(TreeNode *node)
   {
   if (node == nullptr)
   {
   return 0;
   }
   return maxT(robberNode(node->lChild) + robberNode(node->rChild),
   NotRobberNode(node->lChild) + NotRobberNode(node->rChild));

}
int robberNode(TreeNode *node)
{
if (node == nullptr)
{
return 0;
}
return node->val+ NotRobberNode(node->lChild)+ NotRobberNode(node->rChild);
}
int getRobberMax(TreeNode *root)
{
return maxT(robberNode(root), NotRobberNode(root));
}
int main()
{
int arr[7] = { 3,2,3,-1,3,-1,1};
TreeNode *root = create(arr, 7, 0);
cout << getRobberMax(root) << endl;
system(“pause”);
return 0;
}