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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 number123
.
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:
![Sum Root to Leaf Numbers LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct] 2 num1tree](https://assets.leetcode.com/uploads/2021/02/19/num1tree.jpg)
Input: root = [1,2,3] Output: 25 Explanation: The root-to-leaf path1->2
represents the number12
. The root-to-leaf path1->3
represents the number13
. Therefore, sum = 12 + 13 =25
.
Example 2:
![Sum Root to Leaf Numbers LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct] 3 num2tree](https://assets.leetcode.com/uploads/2021/02/19/num2tree.jpg)
Input: root = [4,9,0,5,1] Output: 1026 Explanation: The root-to-leaf path4->9->5
represents the number 495. The root-to-leaf path4->9->1
represents the number 491. The root-to-leaf path4->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
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