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:

  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.

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:

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++

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

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/