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Total Submissions: 3744
Difficulty: Medium
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.
分析：
以下的答案有错，不知道错在哪里！
！
！难道不是统计偶数层的和与奇数层的和，然后比較大小就可得出结果？
/**
* 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<TreeNode*> que;//用来总是保存当层的节点
que.push(root);
int oddsum =root->val;//用于统计奇数层的和
int evensum=0; //用于统偶数层的和
//获取每一层的节点
int curlevel=2;
while(!que.empty())
{
int levelSize = que.size();//通过size来推断当层的结束
for(int i=0; i<levelSize; i++)
{
if(que.front()->left != NULL) //先获取该节点下一层的左右子，再获取该节点的元素。由于一旦压入必弹出。所以先处理左右子
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;//奇数层的和与偶数层的和谁更大谁就是结果
}
};学习别人的代码：
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);
}
注：本博文为EbowTang原创，兴许可能继续更新本文。假设转载，请务必复制本条信息！
原文地址：http://blog.csdn.net/ebowtang/article/details/50890931
原作者博客：http://blog.csdn.net/ebowtang
本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895
转载于:https://www.cnblogs.com/yutingliuyl/p/7225828.html
发布者：全栈程序员-用户IM，转载请注明出处：https://javaforall.cn/108303.html原文链接：https://javaforall.cn
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