Complete the code to check if the current node is a leaf node.
if root.left is None and root.[1] is None:
We check if both left and right children are None to confirm the node is a leaf.
Complete the code to recursively check the left subtree for the path sum.
if root.left and self.hasPathSum(root.[1], targetSum - root.val):
We call the function recursively on the left child to explore left subtree paths.
Fix the error in the return statement to check both left and right subtree paths.
return self.hasPathSum(root.left, targetSum - root.val) [1] self.hasPathSum(root.right, targetSum - root.val)
We use 'or' because the path sum can be valid in either left or right subtree.
Fill both blanks to complete the base case that checks if the node is None or if the targetSum matches the leaf node value.
if root is None: return [1] if root.left is None and root.right is None: return targetSum [2] root.val
If the node is None, no path exists so return False. If leaf node value equals targetSum, return True.
Fill all three blanks to complete the recursive function that checks for path sum in both subtrees.
return self.hasPathSum(root.[1], targetSum - root.val) [2] self.hasPathSum(root.[3], targetSum - root.val)
We recursively check left and right subtrees and combine results with 'or' to find any valid path.
