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

## Description

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].

## Solutions

DFS.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int[] arr;

public boolean isValidSequence(TreeNode root, int[] arr) {
this.arr = arr;
return dfs(root, 0);
}

private boolean dfs(TreeNode root, int u) {
if (root == null || root.val != arr[u]) {
return false;
}
if (u == arr.length - 1) {
return root.left == null && root.right == null;
}
return dfs(root.left, u + 1) || dfs(root.right, u + 1);
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool isValidSequence(TreeNode* root, vector<int>& arr) {
function<bool(TreeNode*, int)> dfs = [&](TreeNode* root, int u) -> bool {
if (!root || root->val != arr[u]) return false;
if (u == arr.size() - 1) return !root->left && !root->right;
return dfs(root->left, u + 1) || dfs(root->right, u + 1);
};
return dfs(root, 0);
}
};

• # 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]) -> bool:
def dfs(root, u):
if root is None or root.val != arr[u]:
return False
if u == len(arr) - 1:
return root.left is None and root.right is None
return dfs(root.left, u + 1) or dfs(root.right, u + 1)

return dfs(root, 0)


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func isValidSequence(root *TreeNode, arr []int) bool {
var dfs func(root *TreeNode, u int) bool
dfs = func(root *TreeNode, u int) bool {
if root == nil || root.Val != arr[u] {
return false
}
if u == len(arr)-1 {
return root.Left == nil && root.Right == nil
}
return dfs(root.Left, u+1) || dfs(root.Right, u+1)
}
return dfs(root, 0)
}