# Root To Leaf Sum Binary Tree

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
/ \
4   8
/   / \
11  13  4
/  \      \
7    2      1
```

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)