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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 1Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
3 / 4 5 / \ \ 1 3 1Maximum 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; } }
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; } };
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原文地址:http://blog.csdn.net/itismelzp/article/details/51627065