From 564fbff98393bc4a720baf61d343864600a50f90 Mon Sep 17 00:00:00 2001
From: SuperVishal0 <115384297+SuperVishal0@users.noreply.github.com>
Date: Mon, 31 Oct 2022 11:16:39 +0530
Subject: [PATCH] Create Path sum

---
 Path sum | 41 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 41 insertions(+)
 create mode 100644 Path sum

diff --git a/Path sum b/Path sum
new file mode 100644
index 0000000..2a7b283
--- /dev/null
+++ b/Path sum	
@@ -0,0 +1,41 @@
+/**
+////////////////////////////////////////
+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 = 22
+Output: true
+Explanation: The root-to-leaf path with the target sum is shown.
+
+
+//////////////////////////////////////////
+
+ * 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:
+    bool hasPathSum(TreeNode* root, int targetSum) {
+        if(!root) return false; // base case
+        if(!root->left and !root->right){ // when you are at leaf node 
+            if(root->val==targetSum) return true; // check if matches or not
+            else return false;
+        }
+        // recursive case
+        return hasPathSum(root->left,targetSum-root->val) || hasPathSum(root->right,targetSum-root->val);
+    }
+};