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Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

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In this post, you will find the solution for the **Path Sum** **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** – Path Sum– LeetCode Problem

Path Sum– LeetCode Problem

**Problem:**

Given the `root`

of a binary tree and an integer `targetSum`

, return `true`

if the tree has a **root-to-leaf** path such that adding up all the values along the path equals `targetSum`

.

A **leaf** is a node with no children.

**Example 1:**

Input:root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22Output:trueExplanation:The root-to-leaf path with the target sum is shown.

**Example 2:**

Input:root = [1,2,3], targetSum = 5Output:falseExplanation:There two root-to-leaf paths in the tree: (1 --> 2): The sum is 3. (1 --> 3): The sum is 4. There is no root-to-leaf path with sum = 5.

**Example 3:**

Input:root = [], targetSum = 0Output:falseExplanation:Since the tree is empty, there are no root-to-leaf paths.

**Constraints:**

- The number of nodes in the tree is in the range
`[0, 5000]`

. `-1000 <= Node.val <= 1000`

`-1000 <= targetSum <= 1000`

Path Sum – LeetCode Solutions

Path Sum Solution in C++:

class Solution { public: bool hasPathSum(TreeNode* root, int sum) { if (!root) return false; if (root->val == sum && !root->left && !root->right) return true; return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val); } };

Path Sum Solution in Java:

class Solution { public boolean hasPathSum(TreeNode root, int sum) { if (root == null) return false; if (root.val == sum && root.left == null && root.right == null) return true; return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val); } }

Path Sum Solution in Python:

class Solution: def hasPathSum(self, root: TreeNode, sum: int) -> bool: if not root: return False if root.val == sum and not root.left and not root.right: return True return self.hasPathSum(root.left, sum - root.val) or \ self.hasPathSum(root.right, sum - root.val)