[LeetCode] House Robber III

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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:

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5
  3
 / \
2   3
 \   \
  3   1

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:

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3
4
5
    3
   / \
  4   5
 / \   \
1   3   1

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.

输入是一个二叉树，要求求出不具有父子关系的最大节点集合。
解题思路如下
最大值有以下两种可能

  1. 选择当前根节点： 当前节点数值 + 左子树最大值(不选择左子树的根节点) + 右子树最大值(不选择右子树的根节点)
  2. 不选择当前根节点: 左子树最大值 + 右子树最大值

我们可以使用递归的方式实现。每次递归运算返回两个值，一个代表选取当前根节点的最大值，一个代表不选取当前根节点的最大值。最后递归结束后，返回两个数值的最大值。

C++ 实现

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int rob(TreeNode* root) {
        if (root == NULL) return 0;
        const pair<int, int> rob_or_not = robOrNot(root);
        return max(rob_or_not.first, rob_or_not.second);
    }
private:
	
    // pair.first 存储不选取当前根节点的结果
    // pair.second 存储选取当前根节点的结果
    pair<int, int> robOrNot(TreeNode* root)
    {
        if (root == NULL) return {0, 0};
        const pair<int, int> left = robOrNot(root->left);
        const pair<int, int> right = robOrNot(root->right);
        // 选择当前根节点的结果，由三部分组成
        int rob_root = left.second + right.second + root->val;
        // 不选择当前根节点的结果，由两部分组成
        int not_rob_root = max(left.first, left.second) + max(right.first, right.second);
        return {rob_root, not_rob_root};
    }
};
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