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.
Credits:
Special thanks to@dietpepsifor adding this problem and creating all test cases.
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Show Tags Show Similar Problems分析：
下面的答案有錯，不知道錯在哪裡！！！難道不是統計偶數層的和與奇數層的和，然後比較大小就可得出結果？
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int rob(TreeNode* root) {
if(root==NULL)
return 0;
queue que;//用來總是保存當層的節點
que.push(root);
int oddsum =root->val;//用於統計奇數層的和
int evensum=0; //用於統偶數層的和
//獲取每一層的節點
int curlevel=2;
while(!que.empty())
{
int levelSize = que.size();//通過size來判斷當層的結束
for(int i=0; ileft != NULL) //先獲取該節點下一層的左右子，再獲取該節點的元素，因為一旦壓入必彈出，所以先處理左右子
que.push(que.front()->left);
if(que.front()->right != NULL)
que.push(que.front()->right);
if(curlevel % 2 ==1)
oddsum += que.front()->val;
else
evensum += que.front()->val;
que.pop();
}
curlevel++;
}
return oddsum > evensum ? oddsum : evensum;//奇數層的和與偶數層的和誰更大誰就是結果
}
};
學習別人的代碼：
int rob(TreeNode* root) {
int child = 0, childchild = 0;
rob(root, child, childchild);
return max(child, childchild);
}
void rob(TreeNode* root, int &child, int &childchild) {
if(!root) return;
int l1 = 0, l2 = 0, r1 = 0, r2 = 0;
rob(root->left, l1, l2);
rob(root->right, r1, r2);
child = l2 + r2 + root->val;
childchild = max(l1, l2) + max(r1, r2);
}
