Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

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

**Example:**

Given the below binary tree and `sum = 22`

,

5/\48// \1113 4 /\\ 721

return true, as there exist a root-to-leaf path `5->4->11->2`

which sum is 22.

## Solution

/** * Definition for a binary tree node. * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int x) { val = x; } * } */ public class Solution { public bool HasPathSum(TreeNode root, int sum) { //Edge case to detact, whenever root is null, return false if(root == null) return false; //Check if it is a leaf node, and it has val == root.val if(root.left == null && root.right == null) { if(root.val == sum) return true; else return false; } if(HasPathSum(root.left,sum-root.val)) { return true; } if(HasPathSum(root.right,sum-root.val)) { return true; } return false; } }

Time Complexity : O(n)