【Sum Root to Leaf Numbers】cpp

题目：

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

代码:

/**
 * 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 Solution::sumFromRoot2Leaf(root, );
        }
        static int sumFromRoot2Leaf(TreeNode* root, int upperVal)
        {
            if ( !root ) return ;
            if ( !root->left && !root->right ) return upperVal*+root->val;
            return Solution::sumFromRoot2Leaf(root->left, upperVal*+root->val) +
                   Solution::sumFromRoot2Leaf(root->right, upperVal*+root->val);
        }
};

tips：

一开始想自低向上算，后来发现自顶向下算，每次传入上一层的值比较科学。

========================================

第二次过这道题，一次AC了，沿用dfs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)
        {
            int ret = ;
            if ( root ) Solution::dfs(root, , ret);
            return ret;
        }
        static void dfs(TreeNode* root, int sum, int& ret )
        {
            if ( !root->left && !root->right )
            {
                sum = sum* + root->val;
                ret += sum;
                return;
            }
            if ( root->left ) Solution::dfs(root->left, sum*+root->val, ret);
            if ( root->right ) Solution::dfs(root->right, sum*+root->val, ret);
        }
};