False.

If the root node's value is equal to x, return True.

Recursively search for x in the left subtree and the right subtree.

Combine the results of the subtrees using an OR operation to determine if x exists in either subtree.

Code

[{"language":"Java","code":"class Solution {\n    public boolean isPresent(TreeNode root, int x) {\n        // Base case: if the node is null, return false\n        if (root == null) {\n            return false;\n        }\n\n        // If the current node matches the value, return true\n        if (root.val == x) {\n            return true;\n        }\n\n        // Recursively check left and right subtrees\n        boolean leftSearch = isPresent(root.left, x);\n        boolean rightSearch = isPresent(root.right, x);\n\n        // Combine the results of left and right subtrees\n        return leftSearch || rightSearch;\n    }\n\n    // Example usage\n    public static void main(String[] args) {\n        TreeNode root = new TreeNode(8);\n        root.left = new TreeNode(3);\n        root.right = new TreeNode(10);\n        root.left.left = new TreeNode(1);\n        root.left.right = new TreeNode(6);\n        root.right.right = new TreeNode(14);\n\n        Solution sol = new Solution();\n        System.out.println(sol.isPresent(root, 6)); // Output: true\n        System.out.println(sol.isPresent(root, 7)); // Output: false\n    }\n}","lang":"java"},{"language":"Python","code":"class Solution:\n    def isPresent(self, root, x):\n        # Base case: if the node is None, return False\n        if not root:\n            return False\n\n        # If the current node matches the value, return True\n        if root.val == x:\n            return True\n\n        # Recursively check left and right subtrees\n        left_search = self.isPresent(root.left, x)\n        right_search = self.isPresent(root.right, x)\n\n        # Combine the results of left and right subtrees\n        return left_search or right_search\n\n\n# Example usage\nif __name__ == \"__main__\":\n    # Construct binary tree\n    root = Solution.TreeNode(8)\n    root.left = Solution.TreeNode(3)\n    root.right = Solution.TreeNode(10)\n    root.left.left = Solution.TreeNode(1)\n    root.left.right = Solution.TreeNode(6)\n    root.right.right = Solution.TreeNode(14)\n\n    sol = Solution()\n    print(sol.isPresent(root, 6))  # Output: True\n    print(sol.isPresent(root, 7))  # Output: False","lang":"python"}]

Complexity

⏰ Time complexity: O(n), as each node in the binary tree is visited once during the recursive traversal.
🧺 Space complexity: O(h) as the recursion depth corresponds to the height of the tree, where H is the height of the binary tree.

Method 2 - Level order traversal

To implement this method, we perform a Breadth-First Search (BFS) by traversing the binary tree level by level using a queue. At each step, we:

Dequeue the front node and check whether it matches the target value x.
If the value is found, return True.
Otherwise, enqueue the left and right children of the current node (if they exist).
If the queue becomes empty and the value is not found, return False.

Code

[{"language":"Java","code":"class Solution {\n    public boolean isPresent(TreeNode root, int x) {\n        // Base case: if the root is null, the value cannot be present\n        if (root == null) {\n            return false;\n        }\n\n        // Create a queue for BFS\n        Queue<TreeNode> queue = new LinkedList<>();\n        queue.offer(root); // Add root to the queue\n\n        while (!queue.isEmpty()) {\n            // Retrieve and remove the front node from the queue\n            TreeNode current = queue.poll();\n\n            // Check

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