House Robber III

1. Description

2. Solution

• Version 1

代码语言：javascript复制

/**
 * 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) {
        	return 0;
        }
        int sum = 0;
        if(root->left) {
        	sum = sum   rob(root->left->left)   rob(root->left->right);
        }
        if(root->right) {
        	sum = sum   rob(root->right->left)   rob(root->right->right);
        }
        return max(root->val   sum, rob(root->left)   rob(root->right));
    }
};

• Version 2

代码语言：javascript复制

/**
 * 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) {
        unordered_map<TreeNode*, int> m;
        return rob(root, m);
    }

private:
    int rob(TreeNode* root, unordered_map<TreeNode*, int>& m) {
    	if(!root) {
        	return 0;
        }
        if(m.find(root) != m.end()) {
        	return m[root];
        }
        int sum = 0;
        if(root->left) {
        	sum = sum   rob(root->left->left, m)   rob(root->left->right, m);
        }
        if(root->right) {
        	sum = sum   rob(root->right->left, m)   rob(root->right->right, m);
        }
        m[root] = max(root->val   sum, rob(root->left, m)   rob(root->right, m));
        return m[root];
    }
};

version

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