Formatted question description: https://leetcode.ca/all/1430.html

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

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].

Solution 1.

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

// Time: O(N)
// Space: O(H)
class Solution {
    bool dfs(TreeNode* root, vector<int> &A, int i) {
        if (i >= A.size()) return false;
        if (!root || root->val != A[i]) return false;
        if (i == A.size() - 1) return !root->left && !root->right;
        return dfs(root->left, A, i + 1) || dfs(root->right, A, i + 1);
    }
public:
    bool isValidSequence(TreeNode* root, vector<int>& arr) {
        if (!root) return arr.size() == 0;
        return dfs(root, arr, 0);
    }
};

Java

  • /**
 * 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) {
        if (root == null || root.val != arr[0])
            return false;
        int length = arr.length;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        int index = 0;
        while (index < length && !queue.isEmpty()) {
            int num = arr[index];
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (node.val == num) {
                    TreeNode left = node.left, right = node.right;
                    if (left != null)
                        queue.offer(left);
                    if (right != null)
                        queue.offer(right);
                    if (index == length - 1 && left == null && right == null)
                        return true;
                }
            }
            index++;
        }
        return false;
    }
}

  • // OJ: https://leetcode.com/problems/check-if-a-string-is-a-valid-sequence-from-root-to-leaves-path-in-a-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
    bool dfs(TreeNode* root, vector<int> &A, int i) {
        if (i >= A.size()) return false;
        if (!root || root->val != A[i]) return false;
        if (i == A.size() - 1) return !root->left && !root->right;
        return dfs(root->left, A, i + 1) || dfs(root->right, A, i + 1);
    }
public:
    bool isValidSequence(TreeNode* root, vector<int>& arr) {
        if (!root) return arr.size() == 0;
        return dfs(root, arr, 0);
    }
};

  • print("Todo!")

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