LeetCode:House Robber III

House Robber III

Total Accepted: 13977  Total Submissions: 37027  Difficulty: Medium

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:

     3
    / \
   2   3
    \   \ 
     3   1

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

Example 2:

     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.

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思路：

原问题分为：1）偷当前结点；2）不偷当前结点；两个问题

1）时的amount 为4个“孙子”结点的值的和；

2）时的amount为2个“孩子”结点的值的和。

java code:

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

    public int rob(TreeNode root) {
        
        if(root == null) return 0;
        
        int valOfNoRoot = 0; // 不偷当前结点
        int valOfRoot = 0; // 偷当前结点
        
        valOfNoRoot = rob(root.left) + rob(root.right);
        
        if(root.left != null)
            valOfRoot += rob(root.left.left) + rob(root.left.right);
        if(root.right != null)
            valOfRoot += rob(root.right.left) + rob(root.right.right);
        
        return Math.max(valOfRoot + root.val, valOfNoRoot);
        
    }

}

这段代码花了：1133 ms

java代码勉强可以通过，但是c++过不了（c++的时间要求就短一些）。

这是由于递归过程中有大量的重复计算，因此可以将计算过的值保存起来，减少计算次数。

使用HashMap来保存已计算值的java code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
 
    public int rob(TreeNode root) {
        HashMap<TreeNode, Integer> map = new HashMap<TreeNode, Integer>();
        return rob(root, map);
    }
    
    private int rob(TreeNode root, HashMap<TreeNode, Integer> map) {
        if(root == null) return 0;
        
        if(map.containsKey(root)) return map.get(root);
        
        int valOfNoRoot = 0; // 不偷当前结点
        int valOfRoot = 0; // 偷当前结点
        
        valOfNoRoot = rob(root.left, map) + rob(root.right, map);
        
        if(root.left != null)
            valOfRoot += rob(root.left.left, map) + rob(root.left.right, map);
        if(root.right != null)
            valOfRoot += rob(root.right.left, map) + rob(root.right.right, map);
        
        int val = Math.max(valOfRoot + root.val, valOfNoRoot);
        map.put(root, val);
        return val;
    }
}

时间缩短到了： 8 ms

c++ code:

/**
 * 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) {
        
        vector<int> ret = robSub(root);
        return max(ret[0], ret[1]);
    }
    
    vector<int> robSub(TreeNode* root) {
        
        if(!root) return vector<int>(2);
        
        vector<int> left= robSub(root->left);
        vector<int> right= robSub(root->right);
        
        vector<int> ret(2);
        ret[0] = max(left[0], left[1]) + max(right[0], right[1]); // 不偷当前结点
        ret[1] = root->val + left[0] + right[0]; // 偷当前结点
        
        return ret;
    }
};