/**
* 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 sumNumbers(TreeNode* root) {
return dfs(root,0);
}
//关键知道dfs返回的是本节点以后含有的所有路径的和
int dfs(TreeNode* root,int sum){
if(root == nullptr) return 0;
else if(root->left == nullptr && root->right == nullptr) return sum*10+root->val;
else return dfs(root->left,sum*10+root->val)+dfs(root->right,sum*10+root->val);
}
};