What is the best approach for Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree?

Depth-First Search (DFS) is the most effective approach. Traverse the tree from the root while matching each node value with the corresponding element in the array. When you reach the final array element, confirm that the current node is a leaf. This solution runs in O(n) time with O(h) space for the recursion stack.

What data structure is used in Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree?

The core data structure is a binary tree. The solution typically uses recursion or an explicit stack/queue for traversal. DFS uses the call stack to track the path, while BFS implementations use a queue storing nodes and the current index in the array.

What is the time complexity of Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree?

The optimal solution runs in O(n) time where n is the number of nodes in the tree. Each node is visited at most once during traversal. Space complexity is O(h) for recursive DFS due to the call stack, where h is the height of the binary tree.

Description Hints 2 Solution Video 17:32

1430. Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree

Q: Is Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree easy or hard?

The problem is rated Medium because it combines tree traversal with path validation logic. The traversal itself is standard DFS, but correctly handling the array index and ensuring the path ends at a leaf requires careful boundary checks.

Q: Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree Python/Java solution

Python and Java implementations usually rely on recursive DFS. The function compares the current node value with the array element, then recursively checks the left and right subtrees with the next index. The path is valid only if the final array element matches a leaf node. Time complexity is O(n) with O(h) recursion space.

Q: How to solve Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree in O(n)?

Use a DFS traversal starting from the root and track the index in the array. At each node, check whether the node value matches arr[index]. If it does, continue recursively with index + 1. When you reach the last array element, return true only if the node is a leaf. This ensures every node is processed once, giving O(n) time complexity.

Q: Is Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree asked at Google/Amazon/Meta?

Variants of root-to-leaf path validation problems appear frequently in interviews at companies like Amazon, Google, and Meta. Interviewers often use them to test understanding of tree traversal, recursion, and path validation logic using DFS.

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Given a binary tree where each path going from the root to any leaf form a valid sequence, check if a given string is a valid sequence in such binary tree.

We get the given string from the concatenation of an array of integers arr and the concatenation of all values of the nodes along a path results in a sequence in the given binary tree.

Example 1:

Input: root = [0,1,0,0,1,0,null,null,1,0,0], arr = [0,1,0,1]
Output: true
Explanation: 
The path 0 -> 1 -> 0 -> 1 is a valid sequence (green color in the figure). 
Other valid sequences are: 
0 -> 1 -> 1 -> 0 
0 -> 0 -> 0

Example 2:

Input: root = [0,1,0,0,1,0,null,null,1,0,0], arr = [0,0,1]
Output: false 
Explanation: The path 0 -> 0 -> 1 does not exist, therefore it is not even a sequence.

Example 3:

Input: root = [0,1,0,0,1,0,null,null,1,0,0], arr = [0,1,1]
Output: false
Explanation: The path 0 -> 1 -> 1 is a sequence, but it is not a valid sequence.

Constraints:

1 <= arr.length <= 5000
0 <= arr[i] <= 9
Each node's value is between [0 - 9].

Topics

Hints open one at a time

Hint 1

Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers.

Hint 2

When reaching at final position check if it is a leaf node.

17:32

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1class Solution:
  def solve(self):
      pass


Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

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

Input

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

Input

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

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