337.House-Robber-III.md

337. House Robber III

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题目地址

https://leetcode.com/problems/house-robber-iii/

题目描述

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.

代码

Approach #1 DFS + Memoization

/**
 * 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 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;
  }
}

Approach #2 DFS

1. 如果小偷偷了当前结点，那么它的子结点不能偷
2. 如果小偷不偷当前结点，那么子结点可以偷，也可以不偷（取其中较大的即可）

num[0] is the max value while rob this node, num[1] is max value while not rob this value.

public int rob(TreeNode root) {
    if (root == null) return 0;
    return Math.max(robInclude(root), robExclude(root));
}

public int robInclude(TreeNode node) {
    if(node == null) return 0;
    return robExclude(node.left) + robExclude(node.right) + node.val;
}

public int robExclude(TreeNode node) {
    if(node == null) return 0;
    return rob(node.left) + rob(node.right);
}


class Solution {
  public int rob(TreeNode root) {
    int[] num = dfs(root);
    return Math.max(num[0], num[1]);
  }
  
  private int[] dfs(TreeNode node) {
    if (node == null)		return new int[2];
    int[] left = dfs(node.left);
    int[] right = dfs(node.right);
    int[] res = new int[2];
    res[0] = left[1] + right[1] + node.val; // 偷 node, left和right不偷
    res[1] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]); // 不偷 node, 子节点可以偷，也可以不偷
    return res;
  }
}

Approach #3 Improved Confusion

public int rob(TreeNode root) {
    int[] res = robSub(root);
    return Math.max(res[0], res[1]);
}

private int[] robSub(TreeNode root) {
    if (root == null) return new int[2];
    
    int[] left = robSub(root.left);
    int[] right = robSub(root.right);
    int[] res = new int[2];

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