Tree — Path Sum

(Easy)

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.

Solution:

Solution is exactly like this algorithm. Try and understand the iterative part.

https://takeitoutamber.medium.com/bst-sum-root-to-leaf-numbers-a8f64e3d9ba1

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        sum_to_end = 0
        stack = [(root, 0)]
        found = False
        
        while stack and not found:
            
            node, current_value = stack.pop()
            
            if node:
                current_value = current_value + node.val
                if not node.left and not node.right:
                    if current_value == sum:
                        found = True    
                    current_value = 0
else:
                    stack.append((node.left, current_value))
                    stack.append((node.right, current_value))
        
        return found