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# Sum Root to Leaf Numbers LeetCode Solution

## Problem – Sum Root to Leaf Numbers LeetCode Solution

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.

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`.

## Sum Root to Leaf Numbers LeetCode Solution in Java

``````public int sumNumbers(TreeNode root) {
return sum(root, 0);
}

public int sum(TreeNode n, int s){
if (n == null) return 0;
if (n.right == null && n.left == null) return s*10 + n.val;
return sum(n.left, s*10 + n.val) + sum(n.right, s*10 + n.val);
}
``````

## Sum Root to Leaf Numbers LeetCode Solution in C++

``````class Solution {
public:
int sumNumbers(TreeNode* root, int num=0) {
return root->left == root->right ? num * 10 + root->val :
((root->left ? sumNumbers(root->left, num * 10 + root->val) : 0) +
(root->right ? sumNumbers(root->right, num * 10 + root->val) : 0));
}
};
``````

## Sum Root to Leaf Numbers LeetCode Solution in Python

``````def sumNumbers(self, root):
"""
:type root: TreeNode
:rtype: int
"""
def helper(sum, root):
if not root:
return 0
sum = sum * 10 + root.val
if not root.left and not root.right:
return sum
return helper(sum, root.left) + helper(sum, root.right)

return helper(0, root)
``````
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