House raiding 3

subject ：

The thief found a new area to steal . There is only one entrance to the area , We call it root .

except root outside , Each house has and only has one “ Father “ The house is connected to it . After some reconnaissance , The clever thief realized that “ All the houses in this place are arranged like a binary tree ”. If Two directly connected houses were robbed the same night , The house will alarm automatically .

Given a binary tree root . return Without triggering the alarm , The maximum amount a thief can steal .

Example 1:

Input : root = [3,2,3,null,3,null,1]
Output : 7
explain : The maximum amount a thief can steal in a night 3 + 3 + 1 = 7

Example 2：

Input : root = [3,4,5,1,3,null,1]
Output : 9
explain : The maximum amount a thief can steal in a night 4 + 5 = 9

Code ：

recursive

/**
 * 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 {
    public int rob(TreeNode root) {
        // if(root == null){
        //     return 0;
        // }
        // if(root.left == null &&  root.right == null){
        //     return root.val;
        // }
        // int val1 = root.val;
        // // Consider parent node 
        // if(root.left != null){
        //     val1 += rob(root.left.left) + rob(root.left.right); 
        // }
        // if(root.right != null){
        //     val1 += rob(root.right.left) + rob(root.right.right);
        // }
        // // Ignore parent node 
        // int val2 = rob(root.left) + rob(root.right);
        // return Math.max(val1, val2);

        Map<TreeNode, Integer> map = new HashMap<>();
        return roll(root, map);
    }
    public int roll(TreeNode root, Map<TreeNode, Integer> map){
        if(root == null){
            return 0;
        }
        if(root.left == null && root.right == null){
            return root.val;
        }
        if(map.containsKey(root)){
            return map.get(root);
        }
        int val1 = root.val;
        if(root.left != null){
            val1 += roll(root.left.left, map) + roll(root.left.right, map);
        }
        if(root.right != null){
            val1 += roll(root.right.left, map) + roll(root.right.right, map);
        }
        int val2 = roll(root.left, map) + roll(root.right, map);
        int res = Math.max(val1, val2);
        map.put(root, res);
        return res;
    }
}

Dynamic programming

Refer to someone else's code

class Solution {
    public int rob(TreeNode root) {
        int[] res = roll(root);
        return Math.max(res[0], res[1]);
    }
    
    public int[] roll(TreeNode root){
        int[] res = new int[2];
        if(root == null){
            return res;
        }
        
        int[] left = roll(root.left);
        int[] right = roll(root.right);

        res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        res[1] = root.val + left[0] + right[0];

        return res;
    }
}