leetcode337

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int Rob(TreeNode root)
        {
            int[] num = dfs(root);
            return Math.Max(num[0], num[1]);
        }

        private int[] dfs(TreeNode x)
        {
            if (x == null) return new int[2];
            int[] left = dfs(x.left);
            int[] right = dfs(x.right);
            int[] res = new int[2];
            res[0] = left[1] + right[1] + x.val;
            res[1] = Math.Max(left[0], left[1]) + Math.Max(right[0], right[1]);
            return res;
        }
}

https://leetcode.com/problems/house-robber-iii/#/description

补充一个python的实现：

class Solution:
    def dfs(self,root):
        if root==None:
            return [0,0]#空节点
        counter = [0,0]
        left = self.dfs(root.left)
        right = self.dfs(root.right)

        #取当前节点，则其左右子节点都不能取
        counter[0] =  root.val + left[1] + right[1]
        
        #不取当前节点，则左右子节点是否选取要看子节点选择是否更大
        counter[1] = max(left[0],left[1]) + max(right[0],right[1])
        return counter

    def rob(self, root: 'TreeNode') -> 'int':
        result = self.dfs(root)
        #第一位表示取当前节点的累计值，第二位表示不取当前节点的累计值
        return max(result[0],result[1])