129. Sum Root to Leaf Numbers | Leetcode solution

 129. Sum Root to Leaf Numbers

You are given the root of a binary tree containing digits from 0 to 9 only.

Each root-to-leaf path in the tree represents a number.

• For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123.

Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.

A leaf node is a node with no children.

 

Example 1:

Input: root = [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: root = [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.

 

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• 0 <= Node.val <= 9
• The depth of the tree will not exceed 10.

solution:

/**

 * Definition for a binary tree node.

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}

 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}

 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}

 * };

 */

class Solution {

public:

    void find_nums(vector<int>&nums , TreeNode* root , int num)

    {   

        num = num+root->val;

        if(root->left == NULL && root->right == NULL)

        {

            nums.push_back(num);

            return;

        }

        num = num*10;

        if(root->left != NULL) 

            find_nums(nums ,  root->left , num);

        if(root->right != NULL)

            find_nums(nums,  root->right , num);

        

    }

    int sumNumbers(TreeNode* root) {

        

        vector<int>nums;

        int num = 0;

        find_nums(nums , root , num);

        int size = nums.size();

        int sum = 0;

        for(int i =0 ; i <size ; i++)

        {

            sum += nums[i];

        }

        return sum;

        

    }

};