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

https://leetcode.com/problems/check-if-a-string-is-a-valid-sequence-from-root-to-leaves-path-in-a-binary-tree/

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].

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isValidSequence(self, root: TreeNode, arr: List[int], pos=0) -> bool:
        if root == None:
            return True if pos == len(arr) else False
        if pos == len(arr) or root.val != arr[pos]:
            return False
        if root.left and root.right:
            return self.isValidSequence(root.left, arr, pos + 1) or self.isValidSequence(root.right, arr, pos + 1)
        if root.left:
            return self.isValidSequence(root.left, arr, pos + 1)
        elif root.right:
            return self.isValidSequence(root.right, arr, pos + 1)
        return True if pos == len(arr) - 1 else False