18. 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.

Note: A leaf is a node with no children.

Example:
Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.


Example 2:
Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

Solution: (Using Preorder Traversal)

class Solution
{
public:
    void addLeafNumbers(TreeNode *root, vector<int> &v, string num)
    {
        if (root->left == NULL && root->right == NULL)
        {
            int x = stoi(num + to_string(root->val));
            v.push_back(x);
            return;
        }

        if (root->left)
        {
            addLeafNumbers(root->left, v, num + to_string(root->val));
        }

        if (root->right)
        {
            addLeafNumbers(root->right, v, num + to_string(root->val));
        }

        return;
    }

    int sumNumbers(TreeNode *root)
    {
        if (root == NULL)
        {
            return 0;
        }
        vector<int> v;
        string s = "";
        addLeafNumbers(root, v, s);
        int sum = 0;

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

        return sum;
    }
};

Using an extra space Complexity: O(N) N = Number of leaf nodes

Solution: Optimizing space complexity

class Solution
{
public:
    void addLeafNumbers(TreeNode *root, string num, int &sum)
    {
        if (root->left == NULL && root->right == NULL)
        {
            int x = stoi(num + to_string(root->val));
            sum = sum + x;
            return;
        }

        if (root->left)
        {
            addLeafNumbers(root->left, num + to_string(root->val), sum);
        }

        if (root->right)
        {
            addLeafNumbers(root->right, num + to_string(root->val), sum);
        }

        return;
    }

    int sumNumbers(TreeNode *root)
    {
        if (root == NULL)
        {
            return 0;
        }

        string s = "";
        int sum = 0;

        addLeafNumbers(root, s, sum);

        return sum;
    }
};