How do you solve Sum of Root To Leaf Binary Numbers on LeetCode?

Sum of Root To Leaf Binary Numbers (LeetCode #1022) 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: Binary numbers can be efficiently calculated using bit manipulation.

What is the brute force solution for Sum of Root To Leaf Binary Numbers?

The brute force approach for Sum of Root To Leaf Binary Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. We can find all root-to-leaf paths in the binary tree and convert each path from binary to decimal. This method is straightforward but inefficient as it requires multiple traversals. The optimal Optimal Solution approach improves this to O(n).

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

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

What are the interview tips for Sum of Root To Leaf Binary Numbers?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as trees with only one node or all left/right children. Explain your thought process and approach as you code, as this shows your understanding.

What are common mistakes when solving Sum of Root To Leaf Binary Numbers?

Not handling the case of null nodes properly, which can lead to null pointer exceptions. Forgetting to return the sum from leaf nodes.

#1022

Sum of Root To Leaf Binary Numbers

Easy

TreeDepth-First SearchBinary TreeDepth-First SearchBinary Tree 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)

We can use a depth-first search (DFS) approach while maintaining the current binary number as we traverse the tree. This avoids the need for multiple traversals and directly computes the sum.

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Algorithm

3 steps

1Step 1: Use DFS to traverse the tree, passing the current binary number formed by the path.
2Step 2: For each node, update the current number by shifting left and adding the node's value.
3Step 3: When a leaf node is reached, add the current number to the total sum.

solution.py18 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def sumNumbers(self, root: TreeNode) -> int:
10        def dfs(node, current_number):
11            if not node:
12                return 0
13            current_number = (current_number << 1) | node.val
14            if not node.left and not node.right:
15                return current_number
16            return dfs(node.left, current_number) + dfs(node.right, current_number)
17
18        return dfs(root, 0)

ℹ

Complexity note: The time complexity is O(n) because 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.

1Binary numbers can be efficiently calculated using bit manipulation.
2Depth-first search is a natural fit for tree traversal problems.

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