LeetCode-337. House Robber III
原创
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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:
Input: [3,2,3,null,3,null,1]
3
/ \
2 3
\ \
3 1
Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
Input: [3,4,5,1,3,null,1]
3
/ \
4 5
/ \ \
1 3 1
Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.
题解：先把非叶节点补全，减少情况。
class Solution {
public:
void toFullTree (TreeNode *t) {
if (t != NULL) {
if (t->left != NULL && t->right == NULL) {
t->right = new TreeNode(0);
}
if (t->left == NULL && t->right != NULL) {
t->left = new TreeNode(0);
}
toFullTree(t->left);
toFullTree(t->right);
}
}
void postDP (TreeNode *t) {
if (t != NULL) {
postDP(t->left);
postDP(t->right);
if (t->left != NULL) {
if (t->left->left == NULL && t->right->left == NULL) {
t->val = max(t->val, t->left->val + t->right->val);
}
if (t->left->left != NULL && t->right->left != NULL) {
t->val = max(t->left->val + t->right->val, t->val + t->left->left->val + t->left->right->val + t->right->left->val + t->right->right->val);
}
if (t->left->left != NULL && t->right->left == NULL) {
t->val = max(t->left->val + t->right->val, t->val + t->left->left->val + t->left->right->val);
}
if (t->left->left == NULL && t->right->left != NULL) {
t->val = max(t->left->val + t->right->val, t->val + t->right->left->val + t->right->right->val);
}
}
}
}
int rob(TreeNode* root) {
if (root == NULL) {
return 0;
}
toFullTree(root);
postDP(root);
return root->val;
}
};
