原题链接：Leecode 129. 求根节点到叶节点数字之和



代码一

/**
 * 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:
    vector<int> path;
    int res=0;
    void dfs(TreeNode* root)
    {
        if(root->left==nullptr && root->right==nullptr)
        {
            int tmp=0;
            for(int i=0;i<path.size();i++)
            {
                tmp=tmp*10+path[i];
            }
            res+=tmp;
            return ;
        }
        if(root->left)
        {
            path.push_back(root->left->val);
            dfs(root->left);
            path.pop_back();
        }
        if(root->right)
        {
            path.push_back(root->right->val);
            dfs(root->right);
            path.pop_back();
        }

    }
    int sumNumbers(TreeNode* root) {
        path.push_back(root->val);
        dfs(root);
        return res;
    }
};

代码二：

class Solution {
public:
    int dfs(TreeNode* root,int pre)
    {
        if(root==nullptr) return 0;
        int sum=pre*10+root->val;
        if(root->left==nullptr && root->right==nullptr)  return sum;
        else return dfs(root->left,sum)+dfs(root->right,sum);
    }
    int sumNumbers(TreeNode* root) {
        return dfs(root,0);
    }
};