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In this post, you will find the solution for the **Sum Root to Leaf Numbers** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Sum Root to Leaf Numbers– LeetCode Problem

Sum Root to Leaf Numbers– LeetCode Problem

**Problem:**

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:25Explanation: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:1026Explanation: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`

.

Sum Root to Leaf Numbers– LeetCode Solutions

Sum Root to Leaf Numbers Solution in C++:

class Solution { public: int sumNumbers(TreeNode* root) { int ans = 0; dfs(root, 0, ans); return ans; } private: void dfs(TreeNode* root, int path, int& ans) { if (!root) return; if (!root->left && !root->right) { ans += path * 10 + root->val; return; } dfs(root->left, path * 10 + root->val, ans); dfs(root->right, path * 10 + root->val, ans); } };

Sum Root to Leaf Numbers Solution in Java:

class Solution { public int sumNumbers(TreeNode root) { dfs(root, 0); return ans; } private int ans = 0; private void dfs(TreeNode root, int path) { if (root == null) return; if (root.left == null && root.right == null) { ans += path * 10 + root.val; return; } dfs(root.left, path * 10 + root.val); dfs(root.right, path * 10 + root.val); } }

Sum Root to Leaf Numbers Solution in Python:

class Solution: def sumNumbers(self, root: Optional[TreeNode]) -> int: self.ans = 0 def dfs(root: Optional[TreeNode], path: int) -> None: if not root: return if not root.left and not root.right: self.ans += path * 10 + root.val return dfs(root.left, path * 10 + root.val) dfs(root.right, path * 10 + root.val) dfs(root, 0) return self.ans