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leetcode House Robber III

leetcode 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:

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  3
/ \
2 3
\ \
3 1

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

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    3
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4 5
/ \ \
1 3 1

Maximum amount of money the thief can rob = 4 + 5 = 9. 题目地址:leetcode House Robber III

题意:给定一棵二叉树,求能获取的权值最大和(相邻的不能同时取)

思路: 树形dp

显然有:

  • rob_root = max(rob_L + rob_R , no_rob_L + no_nob_R + root.val)
  • no_rob_root = rob_L + rob_R

递归即可

Python:

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class Solution(object):
def rob(self, root):
"""
:type root: TreeNode
:rtype: int
"""
return self.dfs_rob(root)[0]

def dfs_rob(self, root):
if not root: return 0,0
rob_L, no_rob_L = self.dfs_rob(root.left)
rob_R, no_rob_R = self.dfs_rob(root.right)
return max(no_rob_L + no_rob_R + root.val , rob_L + rob_R), rob_L + rob_R

C++

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class Solution {
public:
int rob(TreeNode* root) {
return dfs(root).first;
}

pair<int, int> dfs(TreeNode* root){
pair<int, int> dp = make_pair(0,0);
if(root){
pair<int, int> dp_L = dfs(root->left);
pair<int, int> dp_R = dfs(root->right);
dp.second = dp_L.first + dp_R.first;
dp.first = max(dp.second ,dp_L.second + dp_R.second + root->val);
}
return dp;
}
};

Java

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public class Solution {
public int rob(TreeNode root) {
return dfs(root)[0];
}

private int[] dfs(TreeNode root) {
int dp[]={0,0};
if(root != null){
int[] dp_L = dfs(root.left);
int[] dp_R = dfs(root.right);
dp[1] = dp_L[0] + dp_R[0];
dp[0] = Math.max(dp[1] ,dp_L[1] + dp_R[1] + root.val);
}
return dp;
}
}

 

本题是leetcode 337  House Robber III 题解,

更多题解可以查看:https://www.hrwhisper.me/leetcode-algorithm-solution/

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