337. 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:

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.

class Solution {
    public int rob(TreeNode root) {
        if (root == null) return 0;
        
        int leftLevel = 0;
        int leftSubLevel = 0;
        if (root.left != null) {
            leftLevel = rob(root.left);
            leftSubLevel = rob(root.left.left) + rob(root.left.right);
        } 
        
        int rightLevel = 0;
        int rightSubLevel = 0;
        if (root.right != null) {
            rightLevel = rob(root.right);
            rightSubLevel = rob(root.right.left) + rob(root.right.right);
        }
        
        return Math.max((root.val + leftSubLevel + rightSubLevel), leftLevel + rightLevel);
    }
}

巧，巧夺天工。dfs

 

 https://www.cnblogs.com/reboot329/p/6127932.html

方法2：

class Solution {
    public int rob(TreeNode root) {
        int[] res = robRecur(root);
        return Math.max(res[0], res[1]);
    }
    
    public int[] robRecur(TreeNode root) {
        if (root == null) {
            return new int[2];
        }
        
        int[] left = robRecur(root.left);
        int[] right = robRecur(root.right);
        int[] res = new int[2];
        // no rob root
        res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        // rob root
        res[1] = root.val + left[0] + right[0];
        
        return res;
    }
}

 res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);是因为不rob root的值是root的left + root的right，left又有robleft,不robleft。right也一样，所以也是求最大值相加。

 

 https://www.cnblogs.com/zebinlin/p/9809302.html

class Solution {
    public int rob(TreeNode root) {
        return robSub(root, new HashMap<>());
    }

    private int robSub(TreeNode root, Map<TreeNode, Integer> map) {
        if (root == null) return 0;
        if (map.containsKey(root)) return map.get(root);

        int val = 0;

        if (root.left != null) {
            val += robSub(root.left.left, map) + robSub(root.left.right, map);
        }

        if (root.right != null) {
            val += robSub(root.right.left, map) + robSub(root.right.right, map);
        }

        val = Math.max(val + root.val, robSub(root.left, map) + robSub(root.right, map));
        map.put(root, val);

        return val;
    }
}

这种方法是recursive + memo，很大程度缩减了时间