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337. 打家劫舍 III

English Version

题目描述

小偷又发现了一个新的可行窃的地区。这个地区只有一个入口，我们称之为 root 。

除了 root 之外，每栋房子有且只有一个“父“房子与之相连。一番侦察之后，聪明的小偷意识到“这个地方的所有房屋的排列类似于一棵二叉树”。 如果 两个直接相连的房子在同一天晚上被打劫 ，房屋将自动报警。

给定二叉树的 root 。返回 在不触动警报的情况下 ，小偷能够盗取的最高金额 。

 

示例 1:

输入: root = [3,2,3,null,3,null,1]
输出: 7 
解释: 小偷一晚能够盗取的最高金额 3 + 3 + 1 = 7

示例 2:

输入: root = [3,4,5,1,3,null,1]
输出: 9
解释: 小偷一晚能够盗取的最高金额 4 + 5 = 9

 

提示：

• 树的节点数在 [1, 104] 范围内
• 0 <= Node.val <= 104

解法

记忆化搜索。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rob(self, root: TreeNode) -> int:
        @cache
        def dfs(root):
            if root is None:
                return 0
            if root.left is None and root.right is None:
                return root.val
            a = dfs(root.left) + dfs(root.right)
            b = root.val
            if root.left:
                b += dfs(root.left.left) + dfs(root.left.right)
            if root.right:
                b += dfs(root.right.left) + dfs(root.right.right)
            return max(a, b)

        return dfs(root)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Map<TreeNode, Integer> memo;

    public int rob(TreeNode root) {
        memo = new HashMap<>();
        return dfs(root);
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        if (memo.containsKey(root)) {
            return memo.get(root);
        }
        int a = dfs(root.left) + dfs(root.right);
        int b = root.val;
        if (root.left != null) {
            b += dfs(root.left.left) + dfs(root.left.right);
        }
        if (root.right != null) {
            b += dfs(root.right.left) + dfs(root.right.right);
        }
        int res = Math.max(a, b);
        memo.put(root, res);
        return res;
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    unordered_map<TreeNode*, int> memo;

    int rob(TreeNode* root) {
        return dfs(root);
    }

    int dfs(TreeNode* root) {
        if (!root) return 0;
        if (memo.count(root)) return memo[root];
        int a = dfs(root->left) + dfs(root->right);
        int b = root->val;
        if (root->left) b += dfs(root->left->left) + dfs(root->left->right);
        if (root->right) b += dfs(root->right->left) + dfs(root->right->right);
        int res = max(a, b);
        memo[root] = res;
        return res;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func rob(root *TreeNode) int {
	memo := make(map[*TreeNode]int)
	var dfs func(root *TreeNode) int
	dfs = func(root *TreeNode) int {
		if root == nil {
			return 0
		}
		if _, ok := memo[root]; ok {
			return memo[root]
		}
		a := dfs(root.Left) + dfs(root.Right)
		b := root.Val
		if root.Left != nil {
			b += dfs(root.Left.Left) + dfs(root.Left.Right)
		}
		if root.Right != nil {
			b += dfs(root.Right.Left) + dfs(root.Right.Right)
		}
		res := max(a, b)
		memo[root] = res
		return res
	}
	return dfs(root)
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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