text: r"""
  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.
  """
  
  
  # Definition for a binary tree node.
  # class TreeNode:
  #     def __init__(self, x):
  #         self.val = x
  #         self.left = None
  #         self.right = None
  
  class Solution:
      def hasPathSum(self, root, sum):
          """DFS"""
          if not root:
              return False
          
          if (not root.left) and (not root.right):
              return root.val == sum
          else:
              return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)
  
      def hasPathSum2(self, root, sum):
          """BFS"""
          if not root:
              return False
          
          cur_level = [root]
          cur_sums = [sum]
          while cur_level:
              next_level = []
              next_sums = []
              for node, cur_sum in zip(cur_level, cur_sums):
                  if not (node.left or node.right):
                      if cur_sum == node.val:
                          return True
                      else:
                          continue
  
                  if node.left:
                      next_level.append(node.left)
                      next_sums.append(cur_sum - node.val)
                  
                  if node.right:
                      next_level.append(node.right)
                      next_sums.append(cur_sum - node.val)
              
              cur_level = next_level
              cur_sums = next_sums
  
          return False
  