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[leetcode] problem 337 – house robber iii

admin 11月 13, 2020 0

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

No.1

Input: [3,2,3,null,3,null,1]

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4
5
 
  3
 / 
2   3
     
  3   1

Output: 7

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

No.2

Input: [3,4,5,1,3,null,1]

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2
3
4
5
 
    3
   / 
  4   5
 /     
1   3   1

Output: 9

Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

Code

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public class  {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}
 
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public int rob(TreeNode root) {
    int[] result = helper(root);
    return Math.max(result[0], result[1]);
}

private int[] helper(TreeNode root) {
    if (root == null)
        return new int[] {0, 0};

    int[] left = helper(root.left);
    int[] right = helper(root.right);

    int[] res = new int[2];
    res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
    res[1] = left[0] + right[0] + root.val;
    return res;
}
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