Sum Root to Leaf Numbers
You are given the root of a binary tree containing digits from 0 to 9 only.
Each root-to-leaf path in the tree represents a number.
- For example, the root-to-leaf path
1 -> 2 -> 3represents the number123.
Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.
A leaf node is a node with no children.
Example 1:
Input: root = [1,2,3] Output: 25 Explanation: The root-to-leaf path1->2represents the number12. The root-to-leaf path1->3represents the number13. Therefore, sum = 12 + 13 =25.
Example 2:
Input: root = [4,9,0,5,1] Output: 1026 Explanation: The root-to-leaf path4->9->5represents the number 495. The root-to-leaf path4->9->1represents the number 491. The root-to-leaf path4->0represents the number 40. Therefore, sum = 495 + 491 + 40 =1026.
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]. 0 <= Node.val <= 9- The depth of the tree will not exceed
10.
# 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: Optional[TreeNode]) -> int:
paths = [(root, root.val)]
i = 0
total = 0
while i < len(paths):
curr, val = paths[i]
if curr:
if not curr.left and not curr.right:
total += val
if curr.left:
paths.append((curr.left, 10 * val + curr.left.val))
if curr.right:
paths.append((curr.right, 10 * val + curr.right.val))
i += 1
return total
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