Is Sum of Root To Leaf Binary Numbers easy or hard?

This problem is classified as Easy. The main challenge is recognizing that each root-to-leaf path forms a binary number and can be built incrementally using bit shifts during DFS traversal. Once that insight is clear, the implementation is straightforward.

Sum of Root To Leaf Binary Numbers Python/Java solution

Both Python and Java implementations use DFS traversal. Maintain an integer representing the current binary number and update it using a left shift and addition at each node. When reaching a leaf, add the value to the total sum. The logic runs in O(n) time and O(h) space.

How to solve Sum of Root To Leaf Binary Numbers in O(n)?

Traverse the tree using DFS while maintaining the binary value formed along the path. At each node, compute the new value using (current << 1) | node.val. When a leaf node is reached, add the computed number to the result. Since every node is visited once, the algorithm runs in O(n) time.

What is the best approach for Sum of Root To Leaf Binary Numbers?

Depth-First Search (DFS) with incremental binary construction is the most efficient approach. While traversing from root to leaf, shift the current value left by one bit and add the node’s bit. Each node is processed once, giving O(n) time complexity and O(h) space from recursion depth.

Is Sum of Root To Leaf Binary Numbers asked at Google/Amazon/Meta?

Tree traversal and DFS path problems like this frequently appear in interviews at companies such as Amazon, Google, and Meta. Variations involving root-to-leaf paths, path sums, or binary interpretation are common in coding interviews focused on tree algorithms.

What data structure is used in Sum of Root To Leaf Binary Numbers?

The core data structure is a binary tree. The algorithm uses Depth-First Search traversal implemented either through recursion (call stack) or an explicit stack for iterative DFS. Bit manipulation is used to efficiently build binary numbers along each path.

What is the time complexity of Sum of Root To Leaf Binary Numbers?

The time complexity is O(n), where n is the number of nodes in the binary tree. Each node is visited exactly once during the DFS traversal. Space complexity is O(h), where h is the height of the tree due to recursion stack or the explicit stack in the iterative solution.

Description Hints 1 Solution Video 15:00

1022. Sum of Root To Leaf Binary Numbers

Q: What is the best approach for Sum of Root To Leaf Binary Numbers?

Depth-First Search (DFS) with incremental binary construction is the most efficient approach. While traversing from root to leaf, shift the current value left by one bit and add the node’s bit. Each node is processed once, giving O(n) time complexity and O(h) space from recursion depth.

Q: Is Sum of Root To Leaf Binary Numbers asked at Google/Amazon/Meta?

Tree traversal and DFS path problems like this frequently appear in interviews at companies such as Amazon, Google, and Meta. Variations involving root-to-leaf paths, path sums, or binary interpretation are common in coding interviews focused on tree algorithms.

Q: What data structure is used in Sum of Root To Leaf Binary Numbers?

The core data structure is a binary tree. The algorithm uses Depth-First Search traversal implemented either through recursion (call stack) or an explicit stack for iterative DFS. Bit manipulation is used to efficiently build binary numbers along each path.

Q: What is the time complexity of Sum of Root To Leaf Binary Numbers?

The time complexity is O(n), where n is the number of nodes in the binary tree. Each node is visited exactly once during the DFS traversal. Space complexity is O(h), where h is the height of the tree due to recursion stack or the explicit stack in the iterative solution.

Easy76.6% acceptance6 companies

You are given the root of a binary tree where each node has a value 0 or 1. Each root-to-leaf path represents a binary number starting with the most significant bit.

For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.

For all leaves in the tree, consider the numbers represented by the path from the root to that leaf. Return the sum of these numbers.

The test cases are generated so that the answer fits in a 32-bits integer.

Example 1:

Input: root = [1,0,1,0,1,0,1]
Output: 22
Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22

Example 2:

Input: root = [0]
Output: 0

Constraints:

The number of nodes in the tree is in the range [1, 1000].
Node.val is 0 or 1.

Asked at 6 companies

Lime Bloomberg Amazon

Google Microsoft Meta

Topics

Hints open one at a time

Hint 1

Find each path, then transform that path to an integer in base 10.

15:00

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

LanguageLoading editor…

1# Definition for a binary tree node.
2# class TreeNode:
3#     def __init__(self, val=0, left=None, right=None):
4#         self.val = val
5#         self.left = left
6#         self.right = right
7class Solution:
8    def sumRootToLeaf(self, root: Optional[TreeNode]) -> int:
9        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[1,0,1,0,1,0,1]

Input

[0]

Your code is saved on this device.Solution