思路：
1. split 两个tree， 他的sum=左+右
2. 那我们可以先求出总和，然后post order 减少重复的便利
3. post order 属于dp 的一种


代码：

/**
 * 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:
    int sum=0;
    long res=0;
    int maxProduct(TreeNode* root) {
        getsum(root);
        dfs(root);
        
        res%=1000000007;
        return res;
    }
    
    void getsum(TreeNode* root){
        if(root==NULL)return;
        sum+=(root->val);
        getsum(root->left);
        getsum(root->right);
    }
    
    
    long dfs(TreeNode* root){
        if(root==NULL)return 0;
        
        long l=dfs(root->left);
        long r=dfs(root->right);
        
        long cur=l+r+root->val;
        long other=sum-cur;
        
        long product=cur*other;
        res=max(res,product);
        
        return cur;
    }
    
    
};