84：Sum Root to Leaf Numbers

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.

解析1：遍历出树的所有路径，然后将这些路径的和相加，该解析方法和 path sum ii 那道题目解析2方法很像(http://blog.youkuaiyun.com/weishenmetlc/article/details/61209222)，因为都要遍历出所有路径， 代码如下：

// 递归法，时间复杂度 O(n)，空间复杂度 O(logn)
class Solution {
public:
        int sumNumbers(TreeNode* root) {
                int sum = 0;
                int tmp = 0;
                sumNumbers(root, tmp, sum);
                return sum;
        }
private:
        void sumNumbers(TreeNode* root, int& tmp, int& sum) {
                if (!root) return;

                tmp = tmp * 10 + root -> val;
                if (!root -> left && !root -> right)
                        sum += tmp;

                sumNumbers(root -> left, tmp, sum);
                sumNumbers(root -> right, tmp, sum);

                tmp /= 10;
        }
};

解析2：先求出从根节点到其左子树的所有路径和，再求出从根节点到其右子树的所有路径和，相比如解析1， 该方法更易理解，代码如下：

下面代码的思想及编写参考了网址https://github.com/soulmachine/leetcode#leetcode题解题目

// 递归法，时间复杂度 O(n)，空间复杂度 O(logn)
class Solution {
public:
        int sumNumbers(TreeNode* root) {
                return dfs(root, 0);
        }
private:
        int dfs(TreeNode* root, int sum) {
                if (!root) return 0;
                if (!root -> left && !root -> right)
                        return 10 * sum + root -> val;

                return dfs(root -> left, sum * 10 + root -> val) + 
                dfs(root -> right, sum * 10 + root -> val);
        }
};