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原题题目

代码实现（首刷自解超时 自顶至底重复太多次）

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     struct TreeNode *left; *     struct TreeNode *right; * }; */#define MAX 9999999999long modulo;long dvalue;long maxmult;long visit(struct TreeNode* root){
       if(!root)        return 0;    return  visit(root->left) + visit(root->right) + root->val;}   void multvisit(struct TreeNode* root,int sum){
       if(root)    {
           long temp = visit(root);        if(fabs(sum - temp - temp) < dvalue)        {
               dvalue = fabs(sum - temp - temp);                 maxmult = (sum - temp) * temp;          }        multvisit(root->left,sum);        multvisit(root->right,sum);    }}long maxProduct(struct TreeNode* root){
       dvalue = MAX;    maxmult = 0;    modulo = pow(10,9) + 7;    long sum = visit(root);    multvisit(root,sum);    return maxmult % modulo;}

代码实现（自底至顶 一遍即可）

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     struct TreeNode *left; *     struct TreeNode *right; * }; */#define MAX 9999999999long modulo;long dvalue;long maxmult;long visit(struct TreeNode* root){
       if(!root)        return 0;    return  visit(root->left) + visit(root->right) + root->val;}   int multvisit(struct TreeNode* root,int sum){
       if(!root)        return 0;    long rootnumber = root->val + multvisit(root->left,sum) + multvisit(root->right,sum);    int temp = fabs(sum - rootnumber - rootnumber);    if(temp < dvalue)    {
           dvalue = temp;        maxmult = (sum - rootnumber) * rootnumber;    }    return rootnumber;}long maxProduct(struct TreeNode* root){
       dvalue = MAX;    maxmult = 0;    modulo = pow(10,9) + 7;    long sum = visit(root);    multvisit(root,sum);    return maxmult % modulo;}

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