题解 | #二叉树根节点到叶子节点的所有路径和#

 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
private:
    int res = 0;
    int cur_res = 0;
public:
    /**
     * 
     * @param root TreeNode类 
     * @return int整型
     */
    int sumNumbers(TreeNode* root) {
        // write code here
        dfs(root);
        return res;
    }
    
    void dfs(TreeNode* root){
        if(root == nullptr){
            return;
        }

        cur_res = cur_res*10 + root->val;
        if(root->left == nullptr && root->right == nullptr){
            res += cur_res;
        }
        dfs(root->left);
        dfs(root->right);
        cur_res = cur_res/10;
    }
};