How do you solve Sum Root to Leaf Numbers on LeetCode?

Sum Root to Leaf Numbers (LeetCode #129) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: Understanding how to traverse a tree recursively is crucial for solving tree-related problems.

What is the brute force solution for Sum Root to Leaf Numbers?

The brute force approach for Sum Root to Leaf Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves traversing each root-to-leaf path in the binary tree and constructing the corresponding number. We then sum all these numbers to get the final result. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum Root to Leaf Numbers?

Sum Root to Leaf Numbers uses the following concepts: Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Recursion, Tree Traversal.

What are the interview tips for Sum Root to Leaf Numbers?

Always clarify the problem statement and constraints before diving into coding. Think aloud while coding to show your thought process to the interviewer. Consider edge cases, such as trees with only one node or very deep trees.

What are common mistakes when solving Sum Root to Leaf Numbers?

Forgetting to check if a node is null before accessing its properties. Not correctly forming the number as you traverse down the tree.

#129

Sum Root to Leaf Numbers

Medium

TreeDepth-First SearchBinary TreeDepth-First SearchRecursionTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(h)

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Intuition

Time O(n)Space O(h)

The optimal solution uses Depth-First Search (DFS) to traverse the tree while maintaining the current number formed by the path. This avoids unnecessary recalculations and efficiently sums the numbers.

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Algorithm

3 steps

1Step 1: Use a recursive DFS function that takes the current node and the number formed so far.
2Step 2: If the current node is a leaf, return the current number.
3Step 3: Otherwise, recursively call the function for the left and right children, updating the current number.

solution.py16 lines

1class TreeNode:
2    def __init__(self, val=0, left=None, right=None):
3        self.val = val
4        self.left = left
5        self.right = right
6
7class Solution:
8    def sumNumbers(self, root: TreeNode) -> int:
9        def dfs(node, current_number):
10            if not node:
11                return 0
12            current_number = current_number * 10 + node.val
13            if not node.left and not node.right:
14                return current_number
15            return dfs(node.left, current_number) + dfs(node.right, current_number)
16        return dfs(root, 0)

ℹ

Complexity note: The time complexity is O(n) since we visit each node exactly once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1Understanding how to traverse a tree recursively is crucial for solving tree-related problems.
2Recognizing leaf nodes is essential for determining when to sum the current number.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.