30daychallenge (30!) | 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 <= 50000 <= arr[i] <= 9- Each node's value is between [0 - 9].
// duynotes.blogspot.com
/**
* 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 {
public boolean isValidSequence(TreeNode root, int[] arr) {
return DFS(root,0,arr);
}
private boolean DFS(TreeNode root, int i, int[] arr){
if (root == null || i>arr.length-1){
return false;
}
if (root.val != arr[i]){
return false;
}
if (root.left == null && root.right == null && i == arr.length-1 && root.val == arr[i]){
return true;
}
return DFS(root.left,i+1,arr) || DFS(root.right,i+1,arr);
}
}
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