Code & Fun

第60天。

今天的题目是House Robber III

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.
Example 2:

 3
/ \

4 5
/ \
1 3 1

Maximum amount of money the thief can rob = 4 + 5 = 9.

之前有刷过House robber,上次是一道dp的题目,这次看起来有点像是dp的题目,但是其实不是。

我们考虑当前节点root,我们有两种选择:

  • 不偷root:这意味着我们对root的孩子没有限制(即可以偷也可以不偷)。
  • root:这意味着我们不能偷root的孩子。

从上面的分析可以看出,对于一个节点,我们可能需要返回两个值,一个是偷了root所得到的money,一个是不偷root所得到的money.我们可以用一个结构体来表示:

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typedef struct {
int rob;
int norob;
}Ret;

假设我们现在得到了root左孩子和右孩子的Ret了,我们现在要构建root本身的Ret.显然rob = left.norob + right.norob + root->val.然后还有norob,这个很容易就写成norob = left.rob + right.rob,这样写就假定了rob > norob的,在上面的分析中,我们是说我们对root的孩子没有限制,既然没有限制,就可以偷也可以不偷,所以norob = max(left.rob,left.norob) + max(right.rob,right.norob).

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int rob(TreeNode* root) {
Ret ret = robRec(root);
return max(ret.rob,ret.norob);
}
Ret robRec(TreeNode *root) {
Ret ret = {0,0};
if (root == nullptr) return ret;
Ret left = robRec(root->left);
Ret right = robRec(root->right);
ret.rob = left.norob + right.norob + root->val;
ret.norob = max(left.rob,left.norob) + max(right.rob,right.norob);
return ret;
}

dicuss中的解法大都是这个思路,只是写法不同而已,有一个写法比较有趣:

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public int rob(TreeNode root) {
if (root == null) return 0;
return Math.max(robInclude(root), robExclude(root));
}

public int robInclude(TreeNode node) {
if(node == null) return 0;
return robExclude(node.left) + robExclude(node.right) + node.val;
}

public int robExclude(TreeNode node) {
if(node == null) return 0;
return rob(node.left) + rob(node.right);
}

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