Path Sum in Python

Suppose we have one tree and a sum. We have to find one path such that if we follow that path, we will get the sum that will be matched with the given sum. Suppose the tree is like [0,-3,9,-10, null,5] and the sum is 14, then there is a path 0 → 9 → 5

To solve this, we will follow these steps.

  • If the root is null, then return False

  • if left and right subtree are empty, then return true when sum – root.val = 0, otherwise false

  • return solve(root.left, sum – root.val) or solve(root.right, sum – root.val)

Let us see the following implementation to get a better understanding −


 Live Demo

# Definition for a binary tree node.
class TreeNode(object):
   def __init__(self, x): = x
      self.left = None
      self.right = None
def insert(temp,data):
   que = []
   while (len(que)):
      temp = que[0]
      if (not temp.left):
         if data is not None:
            temp.left = TreeNode(data)
            temp.left = TreeNode(0)
      if (not temp.right):
         if data is not None:
            temp.right = TreeNode(data)
            temp.right = TreeNode(0)
def make_tree(elements):
   Tree = TreeNode(elements[0])
   for element in elements[1:]:
      insert(Tree, element)
   return Tree
class Solution(object):
   def hasPathSum(self, root, sum):
      :type root: TreeNode
      :type sum: int
      :rtype: bool
      if not root :
         return False
      if not root.left and not root.right and is not None:
         return sum - == 0
      if is not None:
         return self.hasPathSum(root.left, or self.hasPathSum(root.right,
tree1 = make_tree([0,-3,9,-10,None,5])
ob1 = Solution()
print(ob1.hasPathSum(tree1, 14))


tree1 = make_tree([0,-3,9,-10,None,5])