[Leetcode] 129. Sum Root to Leaf Numbers 解题报告

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.

思路：

递归在通常情况下是先解决子问题，然后再汇总子问题，得到母问题的解。然而该题目略有不同：我们需要先解决母问题，然后再去递归地解决子问题。而递归的边界就是我们遇到了叶子节点。所以每次当我们遇到一个叶子结点的时候，就更新全局结果，否则就递归地解决子问题。代码挺容易理解的。

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) {        if (root == NULL) {            return 0;        }        int ret = 0;        sumNumbers(root, ret, 0);        return ret;    }private:    void sumNumbers(TreeNode* root, int &ret, int parent_sum) { // precondition: root cannot be NULL        int value = parent_sum * 10 + root->val;        if (!root->left && !root->right) {            ret += value;            return;        }        if (root->left) {            sumNumbers(root->left, ret, value);        }        if (root->right) {            sumNumbers(root->right, ret, value);        }    }};