## Question

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 = 22Output:trueExplanation:The root-to-leaf path with the target sum is shown.

**Example 2:**

Input:root = [1,2,3], targetSum = 5Output:falseExplanation:There two root-to-leaf paths in the tree: (1 --> 2): The sum is 3. (1 --> 3): The sum is 4. There is no root-to-leaf path with sum = 5.

**Example 3:**

Input:root = [], targetSum = 0Output:falseExplanation:Since the tree is empty, there are no root-to-leaf paths.

**Constraints:**

- The number of nodes in the tree is in the range
`[0, 5000]`

. `-1000 <= Node.val <= 1000`

`-1000 <= targetSum <= 1000`

## Python Solution

def dfs(root,curr,target): global ans if root==None or ans==True: return if curr+root.val==target and root.left==None and root.right==None: ans = True dfs(root.left,curr+root.val,target) dfs(root.right,curr+root.val,target) class Solution: def hasPathSum(self, root: TreeNode, targetSum: int) -> bool: global ans ans = False dfs(root,0,targetSum) return ans