題目描述

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.

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題目大意

給定一棵二叉樹，不能同時選擇相鄰結點的數值，求可能得到的結點值之和的最大值。

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示例

E1

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.

E2

Input: [3,4,5,1,3,null,1]3/ \4 5/ \ \ 1 3 1Output: 9 Explanation:?Maximum amount of money the thief can rob = 4 + 5 = 9.

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解題思路

遞歸遍歷，計算當前結點以及其孫子結點的數值之和，和該結點的兩個孩子結點之和進行比較，返回較大值。

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復雜度分析

時間復雜度：O(N)

空間復雜度：O(N)

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代碼

/*** 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) {int l = 0, r = 0;return dfs(root, l, r);}int dfs(TreeNode* root, int& l, int& r) {if(root == NULL)return 0;int ll = 0, lr = 0, rl = 0, rr = 0;l = dfs(root->left, ll, lr);r = dfs(root->right, rl, rr);return max(root->val + ll + lr + rl + rr, l + r);} };

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轉載于:https://www.cnblogs.com/heyn1/p/11212240.html

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