LeetCode: 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.
地址：https://oj.leetcode.com/problems/sum-root-to-leaf-numbers/
算法：用递归的方法完成的。首先分别计算左右子树的和，并且把子树中所有叶子节点的深度记录下来，然后根的和=左子树的和+右子树的和+根的值*pow10（叶子节点的深度）。代码：
1 /**
2 * Definition for binary tree
3 * struct TreeNode {
4 * int val;
5 * TreeNode *left;
6 * TreeNode *right;
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8 * };
9 */
10 class Solution {
11 public:
12 int sumNumbers(TreeNode *root) {
13 vector<int> level;
14 if(!root) return 0;
15 return sum(root,level);
16 }
17 int sum(TreeNode *root, vector<int> &level){
18 if(!root) return -1;
19 if(!root->left && !root->right){
20 level.push_back(1);
21 return root->val;
22 }
23 vector<int> left;
24 int left_sum = sum(root->left,left);
25 vector<int> right;
26 int right_sum = sum(root->right,right);
27 int total_sum = 0;
28 if(left_sum >= 0){
29 total_sum += left_sum;
30 vector<int>::iterator it = left.begin();
31 for(; it != left.end(); ++it){
32 total_sum += (root->val * int(pow10(*it)));
33 level.push_back(*it + 1);
34 }
35 }
36 if(right_sum >= 0){
37 total_sum += right_sum;
38 vector<int>::iterator it = right.begin();
39 for(; it != right.end(); ++it){
40 total_sum += (root->val * int(pow10(*it)));
41 level.push_back(*it + 1);
42 }
43 }
44 return total_sum;
45 }
46 };
