337. House Robber III

Description

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:

tree

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

Example 2:

tree

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

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

Solution

Recursion with note, time O(n), space O(n)

robRecur(root)的返回值是rob root和not rob root的max。需要用一个HashMap去辅助记录以避免重复递归。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int rob(TreeNode root) {
        return robRecur(root, new HashMap<>());
    }
    
    public int robRecur(TreeNode root, Map map) {
        if (root == null) {
            return 0;
        }
        
        if (map.containsKey(root)) {
            return map.get(root);
        }
        
        int val = 0;
        if (root.left != null) {
            val += robRecur(root.left.left, map) + robRecur(root.left.right, map);
        }
        
        if (root.right != null) {
            val += robRecur(root.right.left, map) + robRecur(root.right.right, map);
        }
        
        int max = Math.max(root.val + val   // rob root
                          , robRecur(root.left, map) + robRecur(root.right, map));  // no rob root
        map.put(root, max);
        return max;
    }
}

Recursion, time O(n), space O(n)

返回一个int[]以区分rob与否，这样就将一个问题拆分成了不重叠的两个子问题，不会造成重复的递归计算。

返回一个array是java里面好用的方式啊。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
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;
    }
}