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.

Example 1:

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.

Example 2:

Input: [3,4,5,1,3,null,1]

     3
    / \
   4   5
  / \   \ 
 1   3   1

Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.
 

方法1：（遞迴演算法）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int rob(TreeNode root) {
    if(root == null) return 0;
    helper(root);
    return root.val;
}
   
private int helper(TreeNode root){
    if(root == null) return 0;
    int leftsub = helper(root.left);
    int rightsub = helper(root.right);
    int left = root.left == null ? 0 : root.left.val;
    int right = root.right == null ? 0 : root.right.val;
    root.val = Math.max(root.val + leftsub + rightsub, left + right);
    return left + right;
        
    }
}
 

時間複雜度:O(n)

空間複雜度：O(n)

方法2:（另一種遞迴思路）

public int rob(TreeNode root) {
    if (root == null) return 0;
    
    int val = 0;
    
    if (root.left != null) {
        val += rob(root.left.left) + rob(root.left.right);
    }
    
    if (root.right != null) {
        val += rob(root.right.left) + rob(root.right.right);
    }
    
    return Math.max(val + root.val, rob(root.left) + rob(root.right));
}

時間複雜度:O(n)

空間複雜度:O(n)