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[], num[]);
        }

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