题目:
解答:
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int sumNumbers(TreeNode* root) 13 { 14 return sumNumbersHelper(root, 0); 15 } 16 17 int sumNumbersHelper(TreeNode *root, int num) 18 { 19 if (NULL == root) 20 { 21 return 0; 22 } 23 24 num = num * 10 + root->val; 25 26 if (root->left == NULL && root->right == NULL) 27 { 28 return num; 29 } 30 else 31 { 32 return sumNumbersHelper(root->left, num) + sumNumbersHelper(root->right, num); 33 } 34 } 35 };
原文:https://www.cnblogs.com/ocpc/p/12818165.html