root of a binary tree and an integer
true if the tree has a root-to-leaf path such that adding up all the values along the path equals
A leaf is a node with no children.
Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true Explanation: The root-to-leaf path with the target sum is shown.
Input: root = [1,2,3], targetSum = 5 Output: false Explanation: 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.
Input: root = , targetSum = 0 Output: false Explanation: Since the tree is empty, there are no root-to-leaf paths.
- The number of nodes in the tree is in the range
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
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