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/**
 * 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:
    pair<int,int> solve(TreeNode* root){
        if(!root) return {0,0};
        if(!root->left && !root->right){
            return {root->val,0}; //included,excluded
        }
        pair<int,int>l=solve(root->left);
        pair<int,int>r=solve(root->right);
        int inc=root->val+l.second+r.second;
        int exc=max(l.first,l.second) +max(r.first,r.second); //if the current root isnt robbed we must choose the max of left and right valuse because they arent adjacent
        return {inc,exc};
    }

    int rob(TreeNode* root) {
        pair<int,int> res=solve(root);
        return max(res.first,res.second);
    }
};