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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)
}

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