剑指 Offer II 049. 从根节点到叶节点的路径数字之和

dfs水题

/**
 * 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;
    void dfs(TreeNode* root, int num)
    {
        num = num * 10 + root->val;
        if(root->left == nullptr && root->right == nullptr) 
        {
            sum += num;
            return;
        }
        if(root->left)
        {
            dfs(root->left, num);
        }
        if(root->right)
            dfs(root->right, num);
        num = (num - root->val) / 10;
    }


    int sumNumbers(TreeNode* root) {
        dfs(root, sum);
        return sum;

    }
};