Path Sum

Tree DFS

Problem

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.

For 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.

Thought Process

Solution

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        if root is None:
            return 
        if root.left == None and root.right == None and root.val == sum:
                return True
    
        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)
        
#Time: O(n) where 'n' is the total number of leaf nodes
#Space: O(n) because the space is the recursion stack
#and worst case is every node has one child