House Robber III Problem LeetCode

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called 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 form a binary tree. It will automatically contact the police if two directly-linked houses were broken into on the same night.

Given the root of the binary tree, return the maximum amount of money the thief can rob without alerting the police.

Example 1:

Input: root = [3,2,3,null,3,null,1]
Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

Input: root = [3,4,5,1,3,null,1]
Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

 NOTE:

• The number of nodes in the tree is in the range [1, 104].
• 0 <= Node.val <= 104

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Solution:

Source: Java

	/**
	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.
	*/
	public class HouseRobberIII{
	public int rob(TreeNode root) {
	if(root==null){
	return 0;
	}
	int[] res = robHelper(root);
	return Math.max(res[0], res[1]);
	}
	/**
	recursively check if including a root node is maximum profit or excluding it
	*/
	public static int[] robHelper(TreeNode root){
	if(root==null){
	return new int[]{0,0};
	}
	int[] left = robHelper(root.left);
	int[] right = robHelper(root.right);
	int[] res = new int[2];
	//when the root is taken, the grand child of it can be taken
	//res[0] holds the profit when root is taken, and res[1] is when the root is not taken
	res[0] = root.val + left[1] + right[1];
	//left[1] holds the max profit when the left node used its child node and not itself
	//right[1] is defined similarly
	res[1] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
	return res;
	}
	}

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