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 1Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
3 / \ 4 5 / \ \ 1 3 1Maximum amount of money the thief can rob = 4 + 5 = 9.
Frankly I could not come up a smart solution to this at first sight, so that I took credit from https://www.hrwhisper.me/leetcode-house-robber-iii/
I really learnt a lot and got more understanding on DP after understanding it.
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;
}
}
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