LeetCoding Challenge 30/30. Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree

题目

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:

1
2
3

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:

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2
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].

思路

DFS.
注意是否达到了叶子节点。

代码

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def isValidSequence(self, root, arr):
        """
        :type root: TreeNode
        :type arr: List[int]
        :rtype: bool
        """
        return self.dfs(root, arr, 0)
    
    def dfs(self, root, arr, index):
        if root == None or index >= len(arr):
            return False
        if root.left == None and root.right == None and root.val == arr[len(arr)-1]:
            return True
        if index < len(arr) and root.val == arr[index]:
            return self.dfs(root.left, arr, index+1) or self.dfs(root.right, arr, index+1)
        return False