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Sum Root to Leaf Numbers LC #129 Medium DFS · Digit Accumulation
Problem

You are given the root of a binary tree containing digits from 0–9. Each root-to-leaf path represents a number (e.g., 1→2→3 = 123). Return the total sum of all root-to-leaf numbers.

Example 1
Input: root = [1,2,3]
Output: 25
Explanation: Paths: 1→2=12, 1→3=13. Sum=12+13=25.
Example 2
Input: root = [4,9,0,5,1]
Output: 1026
Explanation: Paths: 4→9→5=495, 4→9→1=491, 4→0=40. Sum=495+491+40=1026.
Constraints: 1 ≤ nodes ≤ 1000  |  0 ≤ Node.val ≤ 9  |  tree depth ≤ 10
Try Examples
Custom:
Approach
DFS: curr = curr × 10 + node.val at every step
Accumulate the number digit-by-digit as we go deeper. At a leaf, curr is one complete number — add it to the total. Internal nodes return dfs(left, curr) + dfs(right, curr).
Binary Tree
Number Builder
Waiting for first node…
Current Path
—
Leaf Numbers
No leaf reached yet…
State Variables
node.val
—
curr
0
isLeaf?
—
leafNums
[]
Step Logic
Load an example to begin.
Recursion Stack
Stack is empty
Algorithm Steps
Time
O(n)
Space
O(h)
Why it Works

Multiplying by 10 and adding the next digit is exactly how decimal numbers are formed — curr = curr×10 + digit. At each leaf we have the complete root-to-leaf number. Summing the left and right DFS calls automatically accumulates all leaf contributions. No extra path storage needed.