LeetCode 129: Sum Root to Leaf Numbers — Python Solution

LeetCode 129: Sum Root to Leaf Numbers is a Medium binary tree general problem. This Python walkthrough develops a prefix-number DFS solution, ties every code decision to a concrete invariant, and includes the regression check used before publication.

This independent guide paraphrases the task rather than reproducing LeetCode’s prompt. Use the official page for the exact statement, examples, constraints, and submission runner.

Recognizing the pattern

Tree recursion works when the return value has one precise meaning for every subtree; the parent can then combine child results locally.

For this problem specifically, extend the current number by multiplying by ten and adding the node digit; return it at leaves and sum child results. The invariant worth writing beside the code is: The accumulator is exactly the decimal number represented by the current root-to-node path.

Step-by-step algorithm

1. Identify the input state consumed by sumNumbers(root) and initialize the data required by the prefix-number DFS pattern.
2. Extend the current number by multiplying by ten and adding the node digit; return it at leaves and sum child results.
3. After each update, verify the page’s central invariant: The accumulator is exactly the decimal number represented by the current root-to-node path.
4. Finish only after the boundary behavior is covered: A zero digit still shifts the prefix; an empty tree contributes zero.

Python solution

LeetCode supplies the list, tree, or graph node class referenced by this method. The downloadable test suite includes compatible local node definitions so the implementation can also run outside the judge.

class Solution:
    def sumNumbers(self, root):
        def visit(node, prefix):
            if not node:
                return 0
            value = prefix * 10 + node.val
            if not node.left and not node.right:
                return value
            return visit(node.left, value) + visit(node.right, value)
        return visit(root, 0)

Reading the implementation

The main entry point is sumNumbers(root). The named working state includes value; those variables make the prefix-number DFS state visible instead of hiding it in incidental control flow.

The method expresses the transformation directly without a general traversal loop. Early returns stop as soon as the answer is forced, avoiding work that cannot change the result. In concrete terms, extend the current number by multiplying by ten and adding the node digit; return it at leaves and sum child results.

Correctness argument

Base case. Empty or terminal subproblems return a result that already satisfies the claim for that smallest state.

Inductive step. Extend the current number by multiplying by ten and adding the node digit; return it at leaves and sum child results. Assuming child or smaller states are correct, the current call combines only results allowed by the problem, so the accumulator is exactly the decimal number represented by the current root-to-node path.

Conclusion. Every recursive call reduces the remaining state. The base cases terminate, and induction carries the invariant back to the original input, establishing the returned answer.

Complexity and trade-offs

O(n) time and O(h) recursion space. The auxiliary-space figure excludes the returned output unless the output itself is the structure being built.

An explicit stack avoids recursion-depth limits, but it must carry the same state that recursive call frames provide automatically. That comparison is useful in an interview because it explains why the final implementation is preferable, not merely that it passes.

Regression check

The published implementation belongs to a 100-problem suite that is compiled and exercised behaviorally before deployment.

One reference assertion from that suite is shown below. It targets the normal path while the edge conditions in the next section cover the failure-prone boundaries.

assert Solution().sumNumbers(tree([1,2,3]))==25

Common mistakes and edge cases

• Problem-specific boundary: A zero digit still shifts the prefix; an empty tree contributes zero.
• Pattern-level pitfall: Write the empty-subtree result first and keep returned subtree information separate from any global answer updated at a node.
• Invariant check: after every update, confirm that the accumulator is exactly the decimal number represented by the current root-to-node path.

Interview review checklist

• Explain why prefix-number DFS matches the structure of this input.
• State the invariant in one sentence before tracing code: The accumulator is exactly the decimal number represented by the current root-to-node path.
• Derive O(n) time and O(h) recursion space from how many times each element or state is visited.
• Test the boundary explicitly: A zero digit still shifts the prefix; an empty tree contributes zero.

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