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.

从根节点開始，dfs的思路，事实上也就是postorder（先序遍历），遍历路径上的值每次乘以基数10，过程非常easy，代码例如以下：

class Solution {

public:

    int sum;

    void dfs(TreeNode *root, int pre){

        if (root == NULL)

            return;

        int current = root->val + 10 * pre;

        if (root->left == NULL && root->right == NULL){

            sum += current;

            return;

        }

        if (root->left)

            dfs(root->left, current);

        if (root->right)

            dfs(root->right, current);

    }

    int sumNumbers(TreeNode *root) {

        sum = 0;

        dfs(root, 0);

        return sum;

    }

};