129-sum-root-to-leaf-numbers.py

"""
Problem Link: https://leetcode.com/problems/sum-root-to-leaf-numbers/

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
Note: A leaf is a node with no children.

Example:
Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:
Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
"""
# 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 sumNumbers(self, root: TreeNode) -> int:
        
        def dfs(root, cur_sum=0):
            if not root:
                return 0
            
            if not root.left and not root.right:
                return cur_sum*10 + root.val
            
            return dfs(root.left, cur_sum*10 + root.val) + dfs(root.right, cur_sum*10 + root.val)
            
            
        return dfs(root)
            
class Solution1:
    def sumNumbers(self, root: TreeNode) -> int:
        self.sum = 0
        
        def dfs(root, cur_sum=0):
            if not root:
                return
            
            cur_sum = cur_sum*10 + root.val

            if not root.left and not root.right:
                self.sum += cur_sum
                return
            
            dfs(root.left, cur_sum)
            dfs(root.right, cur_sum)
            
        dfs(root)
        return self.sum