129. Sum Root to Leaf Numbers

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.

所用到的结构体：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */

递归函数如下：

class Solution {public:    int getSum(TreeNode *root, int cur_sum){        // 终止条件        if (root == NULL) return 0;        // 处理叶子节点        if (root->left == NULL && root->right == NULL)            return cur_sum * 10 + root->val;        // 递归过程        return getSum(root->left, cur_sum * 10 + root->val) + getSum(root->right, cur_sum * 10 + root->val);    }    // 129. Sum Root to Leaf Numbers     int sumNumbers(TreeNode* root) {        // 得到root为根的和        return getSum(root, 0);    }}

二叉树具有天然的递归结构，这类问题要想清楚：

1. 递归终止条件（包括根节点为NULL怎么处理，遇到叶子节点怎么处理）；
2. 递归的过程（如何定义一个递归求解过程，把一个大规模的问题缩小到相同套路的子问题，从而缩小规模）；
3. 一旦递归过程想清楚了，就可以假设子问题已经得到解决，进而在此基础上求解更大规模的问题。