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.

解法一：

/**
* 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) {
return dfs(root, 0);
}

int dfs(TreeNode *root, int sum) {
if (root == NULL) return 0;

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

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

解法二：

/**
* 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) {
vector<string> result;
if (root == NULL) return 0;
sumNumbers(root, result, to_string(root->val));

int sum = 0;
for (int i = 0; i < result.size(); i++) {
sum += atoi(result[i].c_str());
}

return sum;
}

void sumNumbers(TreeNode *root, vector<string> &result, string str) {
if (root && !root->left && !root->right) {
result.push_back(str);
return;
}

if (root->left)
sumNumbers(root->left, result, str + to_string(root->left->val));
if (root->right)
sumNumbers(root->right, result, str + to_string(root->right->val));
}
};