Create Path_Sum.cpp

Path_Sum.cpp

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+// Time Complexity: O(N)
+// Auxiliary Space: O(log N), stack space
+
+#include <bits/stdc++.h>
+using namespace std;
+
+#define bool int
+
+/* A binary tree node has data, pointer to left child
+and a pointer to right child */
+class node {
+public:
+	int data;
+	node* left;
+	node* right;
+};
+
+/*
+Given a tree and a sum, return true if there is a path from
+the root down to a leaf, such that adding up all the values
+along the path equals the given sum.
+
+Strategy: subtract the node value from the sum when
+recurring down, and check to see if the sum is 0 when you
+when you reach the leaf node.
+*/
+bool hasPathSum(node* Node, int sum)
+{
+	if (Node == NULL)
+		return 0;
+
+	bool ans = 0;
+
+	int subSum = sum - Node->data;
+
+	/* If we reach a leaf node and sum becomes 0 then return
+	* true*/
+	if (subSum == 0 && Node->left == NULL
+		&& Node->right == NULL)
+		return 1;
+
+	/* otherwise check both subtrees */
+	if (Node->left)
+		ans = ans || hasPathSum(Node->left, subSum);
+	if (Node->right)
+		ans = ans || hasPathSum(Node->right, subSum);
+
+	return ans;
+}
+
+/* UTILITY FUNCTIONS */
+/* Helper function that allocates a new node with the
+given data and NULL left and right pointers. */
+node* newnode(int data)
+{
+	node* Node = new node();
+	Node->data = data;
+	Node->left = NULL;
+	Node->right = NULL;
+
+	return (Node);
+}
+
+// Driver's Code
+int main()
+{
+
+	int sum = 21;
+
+	/* Constructed binary tree is
+				10
+			/ \
+			8 2
+		/ \ /
+		3 5 2
+	*/
+	node* root = newnode(10);
+	root->left = newnode(8);
+	root->right = newnode(2);
+	root->left->left = newnode(3);
+	root->left->right = newnode(5);
+	root->right->left = newnode(2);
+
+	// Function call
+	if (hasPathSum(root, sum))
+		cout << "There is a root-to-leaf path with sum "
+			<< sum;
+	else
+		cout << "There is no root-to-leaf path with sum "
+			<< sum;
+
+	return 0;
+}
+
+