[LeetCode] House Robber III

最新推荐文章于 2020-05-10 21:00:27 发布

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探讨在一棵二叉树表示的特殊区域中，窃贼如何在不触动报警的情况下达到抢劫金额的最大化。通过递归算法解决此问题，并给出了具体的实现代码。

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:

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

Example 2:

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

化成递归问题，如果抢劫root，那么结果位 root->val + rob(root.child.child)

如果不抢劫，那么结果为rob(root.child). 两者中取最大值。

特别注意写递归时要避免重复递归的情形，不然会TLE。

采用备忘录方法，对于这个题，有更简单的写法。

/**
 * 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;
        int l=0;int r=0;
        return recursion(root,l,r);
    }
    int recursion(TreeNode* root,int &l,int &r) //记录以左右子树节点做根的子树大小
    {
        if(!root)
            return 0;
        int ll=0; int lr = 0;int rl = 0;int rr = 0; //记录子节点
        l = recursion(root->left,ll,lr);
        r = recursion(root->right,rl,rr);
        return max(root->val+ll+lr+rl+rr,l+r);
    }
    
};