# Path Sum

## Problem

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.

{% hint style="info" %}
For 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.
{% endhint %}

### Thought Process

![](https://1063826111-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-MGdx41c9p2PMgIHbUTK%2F-MNQAJCZM0tM0iZPBR2k%2F-MNQHPVDxSLXImfGi4AP%2FIMG_348EB560FB01-1.jpeg?alt=media\&token=a1a3f64a-cf76-43b0-9077-3c8c098365de)

## Solution

```
class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        if root is None:
            return 
        if root.left == None and root.right == None and root.val == sum:
                return True
    
        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)
        
#Time: O(n) where 'n' is the total number of leaf nodes
#Space: O(n) because the space is the recursion stack
#and worst case is every node has one child
```