Sum Root to Leaf Numbers

 方法：使用递归的思想

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumNumbers(TreeNode* root) {
        if(root == nullptr)
            return 0;
            
        sum = 0;
        sumNumbers(root, root->left, root->right);
        return sum;
    }
    
    void sumNumbers(TreeNode *root, TreeNode *lt, TreeNode *rt)
    {
        if(lt == nullptr && rt == nullptr)
        {
            sum += root->val;
            return;
        }
        
        if(lt == nullptr)
        {
            rt->val += root->val * 10;
            sumNumbers(rt, rt->left, rt->right);
        }
        else if(rt == nullptr)
        {
            lt->val += root->val * 10;
            sumNumbers(lt, lt->left, lt->right);
        }
        else
        {
            lt->val += root->val * 10;
            rt->val += root->val * 10;
            sumNumbers(rt, rt->left, rt->right);
            sumNumbers(lt, lt->left, lt->right);
        }
    }
    
private:
    int sum;
};

更简单的递归写法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumNumbers(TreeNode* root) {
        return sumNumbers(root, 0);
    }
    
    int sumNumbers(TreeNode *root, int sum)
    {
        if(root == nullptr)
            return 0;
        if(root->left == nullptr && root->right == nullptr)
            return (10 * sum + root->val);
            
        return sumNumbers(root->left, 10 * sum + root->val) + sumNumbers(root->right, 10 * sum + root->val);
    }
};